Class 12 NCERT Solutions
Chapter 2: Inverse Trigonometric Functions
Master the domain, range, and principal value branches with our step-by-step logic.
Exercise 2.1
1. Find the principal value of \(\sin^{-1}\!\left(-\dfrac{1}{2}\right)\).
Solution:
The principal value branch of \(\sin^{-1}\) has range \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). Since \(\sin y=-\dfrac{1}{2}<0\), \(y\) lies in the fourth quadrant (negative angle).
\(\sin^{-1}\!\left(-\dfrac{1}{2}\right)=-\sin^{-1}\dfrac{1}{2}=-\sin^{-1}\sin\dfrac{\pi}{6}=-\dfrac{\pi}{6}\)
The principal value branch of \(\sin^{-1}\) has range \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). Since \(\sin y=-\dfrac{1}{2}<0\), \(y\) lies in the fourth quadrant (negative angle).
\(\sin^{-1}\!\left(-\dfrac{1}{2}\right)=-\sin^{-1}\dfrac{1}{2}=-\sin^{-1}\sin\dfrac{\pi}{6}=-\dfrac{\pi}{6}\)
2. Find the principal value of \(\cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)\).
Solution:
The principal value branch of \(\cos^{-1}\) is \([0,\pi]\). Since \(\cos y=\dfrac{\sqrt{3}}{2}>0\), \(y\) lies in the first quadrant.
\(\cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)=\cos^{-1}\cos\dfrac{\pi}{6}=\dfrac{\pi}{6}\)
The principal value branch of \(\cos^{-1}\) is \([0,\pi]\). Since \(\cos y=\dfrac{\sqrt{3}}{2}>0\), \(y\) lies in the first quadrant.
\(\cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)=\cos^{-1}\cos\dfrac{\pi}{6}=\dfrac{\pi}{6}\)
3. Find the principal value of \(\operatorname{cosec}^{-1}(2)\).
Solution:
Since \(x=2>0\), the value lies in the first quadrant.
\(\theta=\operatorname{cosec}^{-1}2=\operatorname{cosec}^{-1}\operatorname{cosec}\dfrac{\pi}{6}=\dfrac{\pi}{6}\)
Since \(x=2>0\), the value lies in the first quadrant.
\(\theta=\operatorname{cosec}^{-1}2=\operatorname{cosec}^{-1}\operatorname{cosec}\dfrac{\pi}{6}=\dfrac{\pi}{6}\)
4. Find the principal value of \(\tan^{-1}(-\sqrt{3})\).
Solution:
The principal branch of \(\tan^{-1}\) is \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\). Since \(\tan y=-\sqrt{3}<0\), \(y\) is a negative angle in the fourth quadrant.
\(\tan^{-1}(-\sqrt{3})=-\tan^{-1}\sqrt{3}=-\tan^{-1}\tan\dfrac{\pi}{3}=-\dfrac{\pi}{3}\)
The principal branch of \(\tan^{-1}\) is \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\). Since \(\tan y=-\sqrt{3}<0\), \(y\) is a negative angle in the fourth quadrant.
\(\tan^{-1}(-\sqrt{3})=-\tan^{-1}\sqrt{3}=-\tan^{-1}\tan\dfrac{\pi}{3}=-\dfrac{\pi}{3}\)
5. Find the principal value of \(\cos^{-1}\!\left(-\dfrac{1}{2}\right)\).
Solution:
Since \(\cos y=-\dfrac{1}{2}<0\), \(y\) lies in the second quadrant: \(y=\pi-\theta\).
\(\cos^{-1}\!\left(-\dfrac{1}{2}\right)=\pi-\cos^{-1}\dfrac{1}{2}=\pi-\cos^{-1}\cos\dfrac{\pi}{3}=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}\)
Since \(\cos y=-\dfrac{1}{2}<0\), \(y\) lies in the second quadrant: \(y=\pi-\theta\).
\(\cos^{-1}\!\left(-\dfrac{1}{2}\right)=\pi-\cos^{-1}\dfrac{1}{2}=\pi-\cos^{-1}\cos\dfrac{\pi}{3}=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}\)
6. Find the principal value of \(\tan^{-1}(-1)\).
Solution:
Since \(x=-1<0\), the value of \(\tan^{-1}x\) lies between \(-\dfrac{\pi}{2}\) and \(0\).
\(\tan^{-1}(-1)=-\tan^{-1}1=-\tan^{-1}\tan\dfrac{\pi}{4}=-\dfrac{\pi}{4}\)
Since \(x=-1<0\), the value of \(\tan^{-1}x\) lies between \(-\dfrac{\pi}{2}\) and \(0\).
\(\tan^{-1}(-1)=-\tan^{-1}1=-\tan^{-1}\tan\dfrac{\pi}{4}=-\dfrac{\pi}{4}\)
7. Find the principal value of \(\sec^{-1}\!\left(\dfrac{2}{\sqrt{3}}\right)\).
Solution:
The range of \(\sec^{-1}\) is \([0,\pi]-\!\left\{\dfrac{\pi}{2}\right\}\). Since \(\sec y=\dfrac{2}{\sqrt{3}}>0\), \(y\) is in the first quadrant.
\(\sec^{-1}\!\left(\dfrac{2}{\sqrt{3}}\right)=\sec^{-1}\!\left(\sec\dfrac{\pi}{6}\right)=\dfrac{\pi}{6}\)
The range of \(\sec^{-1}\) is \([0,\pi]-\!\left\{\dfrac{\pi}{2}\right\}\). Since \(\sec y=\dfrac{2}{\sqrt{3}}>0\), \(y\) is in the first quadrant.
\(\sec^{-1}\!\left(\dfrac{2}{\sqrt{3}}\right)=\sec^{-1}\!\left(\sec\dfrac{\pi}{6}\right)=\dfrac{\pi}{6}\)
8. Find the principal value of \(\cot^{-1}(\sqrt{3})\).
Solution:
Since \(x=\sqrt{3}>0\), the value lies in the first quadrant.
\(\theta=\cot^{-1}\sqrt{3}=\cot^{-1}\cot\dfrac{\pi}{6}=\dfrac{\pi}{6}\)
Since \(x=\sqrt{3}>0\), the value lies in the first quadrant.
\(\theta=\cot^{-1}\sqrt{3}=\cot^{-1}\cot\dfrac{\pi}{6}=\dfrac{\pi}{6}\)
9. Find the principal value of \(\cos^{-1}\!\left(-\dfrac{1}{\sqrt{2}}\right)\).
Solution:
Since \(x=-\dfrac{1}{\sqrt{2}}<0\), the value of \(\cos^{-1}x\) lies between \(\dfrac{\pi}{2}\) and \(\pi\).
\(\cos^{-1}\!\left(-\dfrac{1}{\sqrt{2}}\right)=\pi-\cos^{-1}\dfrac{1}{\sqrt{2}}=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}\)
Since \(x=-\dfrac{1}{\sqrt{2}}<0\), the value of \(\cos^{-1}x\) lies between \(\dfrac{\pi}{2}\) and \(\pi\).
\(\cos^{-1}\!\left(-\dfrac{1}{\sqrt{2}}\right)=\pi-\cos^{-1}\dfrac{1}{\sqrt{2}}=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}\)
10. Find the principal value of \(\operatorname{cosec}^{-1}(-\sqrt{2})\).
Solution:
Since \(\operatorname{cosec}y=-\sqrt{2}<0\), \(y\) lies in the fourth quadrant (negative angle).
\(\operatorname{cosec}^{-1}(-\sqrt{2})=-\operatorname{cosec}^{-1}\sqrt{2}=-\operatorname{cosec}^{-1}\operatorname{cosec}\dfrac{\pi}{4}=-\dfrac{\pi}{4}\)
Since \(\operatorname{cosec}y=-\sqrt{2}<0\), \(y\) lies in the fourth quadrant (negative angle).
\(\operatorname{cosec}^{-1}(-\sqrt{2})=-\operatorname{cosec}^{-1}\sqrt{2}=-\operatorname{cosec}^{-1}\operatorname{cosec}\dfrac{\pi}{4}=-\dfrac{\pi}{4}\)
11. Find the value of \(\tan^{-1}(1)+\cos^{-1}\!\left(-\dfrac{1}{2}\right)+\sin^{-1}\!\left(-\dfrac{1}{2}\right)\).
Solution:
Substitute the known principal values and simplify:
\(=\tan^{-1}1+\left(\pi-\cos^{-1}\dfrac{1}{2}\right)+\left(-\sin^{-1}\dfrac{1}{2}\right)\)
\(=\dfrac{\pi}{4}+\pi-\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{3\pi+12\pi-4\pi-2\pi}{12}=\dfrac{9\pi}{12}=\dfrac{3\pi}{4}\)
Substitute the known principal values and simplify:
\(=\tan^{-1}1+\left(\pi-\cos^{-1}\dfrac{1}{2}\right)+\left(-\sin^{-1}\dfrac{1}{2}\right)\)
\(=\dfrac{\pi}{4}+\pi-\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{3\pi+12\pi-4\pi-2\pi}{12}=\dfrac{9\pi}{12}=\dfrac{3\pi}{4}\)
12. Find the value of \(\cos^{-1}\!\left(\dfrac{1}{2}\right)+2\sin^{-1}\!\left(\dfrac{1}{2}\right)\).
Solution:
\(\cos^{-1}\cos\dfrac{\pi}{3}+2\sin^{-1}\sin\dfrac{\pi}{6}=\dfrac{\pi}{3}+2\cdot\dfrac{\pi}{6}=\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}\)
\(\cos^{-1}\cos\dfrac{\pi}{3}+2\sin^{-1}\sin\dfrac{\pi}{6}=\dfrac{\pi}{3}+2\cdot\dfrac{\pi}{6}=\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}\)
13. If \(\sin^{-1}x=y\), then: (A) \(0\leq y\leq\pi\) (B) \(-\dfrac{\pi}{2}\leq y\leq\dfrac{\pi}{2}\) (C) \(0
Solution:
By definition of the principal value branch of \(\sin^{-1}\), the range is the closed interval \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). Correct answer: (B)
14. \(\tan^{-1}\sqrt{3}-\sec^{-1}(-2)\) is equal to: (A) \(\pi\) (B) \(-\dfrac{\pi}{3}\) (C) \(\dfrac{\pi}{3}\) (D) \(\dfrac{2\pi}{3}\)
Solution:
Use \(\sec^{-1}(-x)=\pi-\sec^{-1}x\):
\(\tan^{-1}\sqrt{3}-(\pi-\sec^{-1}2)=\dfrac{\pi}{3}-\pi+\dfrac{\pi}{3}=\dfrac{2\pi-3\pi}{3}=-\dfrac{\pi}{3}\)
Correct answer: (B)
Solution:
By definition of the principal value branch of \(\sin^{-1}\), the range is the closed interval \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). Correct answer: (B)
By definition of the principal value branch of \(\sin^{-1}\), the range is the closed interval \(\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\). Correct answer: (B)
14. \(\tan^{-1}\sqrt{3}-\sec^{-1}(-2)\) is equal to: (A) \(\pi\) (B) \(-\dfrac{\pi}{3}\) (C) \(\dfrac{\pi}{3}\) (D) \(\dfrac{2\pi}{3}\)
Solution:
Use \(\sec^{-1}(-x)=\pi-\sec^{-1}x\):
\(\tan^{-1}\sqrt{3}-(\pi-\sec^{-1}2)=\dfrac{\pi}{3}-\pi+\dfrac{\pi}{3}=\dfrac{2\pi-3\pi}{3}=-\dfrac{\pi}{3}\)
Correct answer: (B)
Use \(\sec^{-1}(-x)=\pi-\sec^{-1}x\):
\(\tan^{-1}\sqrt{3}-(\pi-\sec^{-1}2)=\dfrac{\pi}{3}-\pi+\dfrac{\pi}{3}=\dfrac{2\pi-3\pi}{3}=-\dfrac{\pi}{3}\)
Correct answer: (B)
Exercise 2.2
1. Prove that \(3\sin^{-1}x=\sin^{-1}(3x-4x^3),\ x\in\left[-\dfrac{1}{2},\dfrac{1}{2}\right]\).
Solution:
Start from the standard triple angle identity \(\sin3\theta=3\sin\theta-4\sin^3\theta\). Substitute \(\sin\theta=x\) (so \(\theta=\sin^{-1}x\)):
\(\sin3\theta=3x-4x^3\Rightarrow3\theta=\sin^{-1}(3x-4x^3)\)
Replacing \(\theta=\sin^{-1}x\): \(3\sin^{-1}x=\sin^{-1}(3x-4x^3)\).
Start from the standard triple angle identity \(\sin3\theta=3\sin\theta-4\sin^3\theta\). Substitute \(\sin\theta=x\) (so \(\theta=\sin^{-1}x\)):
\(\sin3\theta=3x-4x^3\Rightarrow3\theta=\sin^{-1}(3x-4x^3)\)
Replacing \(\theta=\sin^{-1}x\): \(3\sin^{-1}x=\sin^{-1}(3x-4x^3)\).
2. Prove that \(3\cos^{-1}x=\cos^{-1}(4x^3-3x),\ x\in\left[\dfrac{1}{2},1\right]\).
Solution:
Let \(\cos^{-1}x=\theta\), so \(x=\cos\theta\). Apply the triple angle identity \(\cos3\theta=4\cos^3\theta-3\cos\theta=4x^3-3x\).
\(\Rightarrow3\theta=\cos^{-1}(4x^3-3x)\Rightarrow3\cos^{-1}x=\cos^{-1}(4x^3-3x)\).
Let \(\cos^{-1}x=\theta\), so \(x=\cos\theta\). Apply the triple angle identity \(\cos3\theta=4\cos^3\theta-3\cos\theta=4x^3-3x\).
\(\Rightarrow3\theta=\cos^{-1}(4x^3-3x)\Rightarrow3\cos^{-1}x=\cos^{-1}(4x^3-3x)\).
3. Write \(\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x},\ x\neq0\) in simplest form.
Solution:
Substitute \(x=\tan\theta\) (so \(\theta=\tan^{-1}x\)). Then \(\sqrt{1+x^2}=\sec\theta\).
\(\tan^{-1}\!\left(\dfrac{\sec\theta-1}{\tan\theta}\right)=\tan^{-1}\!\left(\dfrac{1-\cos\theta}{\sin\theta}\right)=\tan^{-1}\!\left(\dfrac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)=\tan^{-1}\!\left(\tan\dfrac{\theta}{2}\right)=\dfrac{\theta}{2}=\dfrac{1}{2}\tan^{-1}x\)
Substitute \(x=\tan\theta\) (so \(\theta=\tan^{-1}x\)). Then \(\sqrt{1+x^2}=\sec\theta\).
\(\tan^{-1}\!\left(\dfrac{\sec\theta-1}{\tan\theta}\right)=\tan^{-1}\!\left(\dfrac{1-\cos\theta}{\sin\theta}\right)=\tan^{-1}\!\left(\dfrac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right)=\tan^{-1}\!\left(\tan\dfrac{\theta}{2}\right)=\dfrac{\theta}{2}=\dfrac{1}{2}\tan^{-1}x\)
4. Write \(\tan^{-1}\!\sqrt{\dfrac{1-\cos x}{1+\cos x}},\ x<\pi\) in simplest form.
Solution:
Use half-angle identities \(1-\cos x=2\sin^2\dfrac{x}{2}\) and \(1+\cos x=2\cos^2\dfrac{x}{2}\):
\(\tan^{-1}\!\sqrt{\dfrac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}=\tan^{-1}\!\left|\tan\dfrac{x}{2}\right|=\tan^{-1}\tan\dfrac{x}{2}=\dfrac{x}{2}\)
Use half-angle identities \(1-\cos x=2\sin^2\dfrac{x}{2}\) and \(1+\cos x=2\cos^2\dfrac{x}{2}\):
\(\tan^{-1}\!\sqrt{\dfrac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}=\tan^{-1}\!\left|\tan\dfrac{x}{2}\right|=\tan^{-1}\tan\dfrac{x}{2}=\dfrac{x}{2}\)
5. Write \(\tan^{-1}\!\left(\dfrac{\cos x-\sin x}{\cos x+\sin x}\right),\ 0
Solution:
Divide numerator and denominator by \(\cos x\), then apply the tangent subtraction formula:
\(=\tan^{-1}\!\left(\dfrac{1-\tan x}{1+\tan x}\right)=\tan^{-1}\!\left(\dfrac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\right)=\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-x\right)=\dfrac{\pi}{4}-x\)
6. Write \(\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}},\ |x|
Solution:
Substitute \(x=a\sin\theta\) (so \(\theta=\sin^{-1}\dfrac{x}{a}\)):
\(\tan^{-1}\!\left(\dfrac{a\sin\theta}{\sqrt{a^2-a^2\sin^2\theta}}\right)=\tan^{-1}\!\left(\dfrac{a\sin\theta}{a\cos\theta}\right)=\tan^{-1}(\tan\theta)=\theta=\sin^{-1}\dfrac{x}{a}\)
7. Write \(\tan^{-1}\!\left(\dfrac{3a^2x-x^3}{a^3-3ax^2}\right),\ a>0,\ -\dfrac{a}{\sqrt{3}}\leq x\leq\dfrac{a}{\sqrt{3}}\) in simplest form.
Solution:
Divide numerator and denominator by \(a^3\), then substitute \(\dfrac{x}{a}=\tan\theta\):
\(\tan^{-1}\!\left(\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\right)=\tan^{-1}(\tan3\theta)=3\theta=3\tan^{-1}\dfrac{x}{a}\)
8. Find the value of \(\tan^{-1}\!\left[2\cos\!\left(2\sin^{-1}\dfrac{1}{2}\right)\right]\).
Solution:
Work from the inside outward, substituting known values step by step:
\(=\tan^{-1}\!\left[2\cos\!\left(2\cdot\dfrac{\pi}{6}\right)\right]=\tan^{-1}\!\left[2\cos\dfrac{\pi}{3}\right]=\tan^{-1}\!\left[2\cdot\dfrac{1}{2}\right]=\tan^{-1}1=\dfrac{\pi}{4}\)
9. Find the value of \(\tan\dfrac{1}{2}\!\left[\sin^{-1}\dfrac{2x}{1+x^2}+\cos^{-1}\dfrac{1-y^2}{1+y^2}\right]\), \(|x|<1,y>0,xy<1\).
Solution:
Substitute \(x=\tan\theta\) and \(y=\tan\phi\). Use the double-angle identities \(\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}\) and \(\cos2\phi=\dfrac{1-\tan^2\phi}{1+\tan^2\phi}\):
\(=\tan\!\left[\dfrac{1}{2}\sin^{-1}(\sin2\theta)+\dfrac{1}{2}\cos^{-1}(\cos2\phi)\right]=\tan(\theta+\phi)=\dfrac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}=\dfrac{x+y}{1-xy}\)
10. Find the value of \(\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)\).
Solution:
Since \(\dfrac{2\pi}{3}\notin\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\), we cannot apply the identity directly. Rewrite using the supplementary angle:
\(\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)=\sin^{-1}\!\left(\sin\!\left(\pi-\dfrac{\pi}{3}\right)\right)=\sin^{-1}\!\left(\sin\dfrac{\pi}{3}\right)=\dfrac{\pi}{3}\)
11. Find the value of \(\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)\).
Solution:
Since \(\dfrac{3\pi}{4}\notin\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), rewrite using the identity \(\tan(\pi-\theta)=-\tan\theta\):
\(\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)=\tan^{-1}\!\left(\tan\!\left(\pi-\dfrac{\pi}{4}\right)\right)=\tan^{-1}\!\left(-\tan\dfrac{\pi}{4}\right)=-\tan^{-1}\tan\dfrac{\pi}{4}=-\dfrac{\pi}{4}\)
12. Find the value of \(\tan\!\left(\sin^{-1}\dfrac{3}{5}+\cot^{-1}\dfrac{3}{2}\right)\).
Solution:
Let \(\sin^{-1}\dfrac{3}{5}=x\) and \(\cot^{-1}\dfrac{3}{2}=y\). Both angles are in the first quadrant.
From \(\sin x=\dfrac{3}{5}\): \(\cos x=\dfrac{4}{5}\), so \(\tan x=\dfrac{3}{4}\).
From \(\cot y=\dfrac{3}{2}\): \(\tan y=\dfrac{2}{3}\).
\(\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\dfrac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot\frac{2}{3}}=\dfrac{\frac{17}{12}}{\frac{1}{2}}=\dfrac{17}{6}\)
13. \(\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)\) is equal to: (A) \(\dfrac{7\pi}{6}\) (B) \(\dfrac{5\pi}{6}\) (C) \(\dfrac{\pi}{3}\) (D) \(\dfrac{\pi}{6}\)
Solution:
Since \(\cos\dfrac{7\pi}{6}<0\), the result must lie in \(\left[\dfrac{\pi}{2},\pi\right]\). Use \(\cos(2\pi-\theta)=\cos\theta\):
\(\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)=\cos^{-1}\!\left(\cos\!\left(2\pi-\dfrac{7\pi}{6}\right)\right)=\dfrac{5\pi}{6}\)
Correct answer: (B)
14. \(\sin\!\left(\dfrac{\pi}{3}-\sin^{-1}\!\left(-\dfrac{1}{2}\right)\right)\) is equal to: (A) \(\dfrac{1}{2}\) (B) \(\dfrac{1}{3}\) (C) \(\dfrac{1}{4}\) (D) 1
Solution:
Use \(\sin^{-1}(-x)=-\sin^{-1}x\) to simplify the inner expression:
\(\sin\!\left(\dfrac{\pi}{3}+\sin^{-1}\dfrac{1}{2}\right)=\sin\!\left(\dfrac{\pi}{3}+\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{2}=1\)
Correct answer: (D)
15. \(\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})\) is equal to: (A) \(\pi\) (B) \(-\dfrac{\pi}{2}\) (C) 0 (D) \(2\sqrt{3}\)
Solution:
Use \(\cot^{-1}(-x)=\pi-\cot^{-1}x\):
\(\tan^{-1}\sqrt{3}-(\pi-\cot^{-1}\sqrt{3})=\dfrac{\pi}{3}-\pi+\dfrac{\pi}{6}=\dfrac{2\pi-6\pi+\pi}{6}=-\dfrac{3\pi}{6}=-\dfrac{\pi}{2}\)
Correct answer: (B)
Miscellaneous Exercise (Chapter 2)
1. Find the value of \(\cos^{-1}\!\left(\cos\dfrac{13\pi}{6}\right)\).
Solution:
Reduce the angle: \(\cos\dfrac{13\pi}{6}=\cos\!\left(2\pi+\dfrac{\pi}{6}\right)=\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}>0\). The result lies in the first quadrant.
\(\cos^{-1}\!\left(\cos\dfrac{13\pi}{6}\right)=\cos^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{6}\)
2. Find the value of \(\tan^{-1}\!\left(\tan\dfrac{7\pi}{6}\right)\).
Solution:
Reduce: \(\tan\dfrac{7\pi}{6}=\tan\!\left(\pi+\dfrac{\pi}{6}\right)=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}>0\). The result lies in the first quadrant.
\(\tan^{-1}\!\left(\tan\dfrac{7\pi}{6}\right)=\tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}\)
3. Prove that \(2\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}\).
Solution:
Let \(\sin^{-1}\dfrac{3}{5}=\theta\) (first quadrant). Then \(\cos\theta=\dfrac{4}{5}\) and \(\tan\theta=\dfrac{3}{4}\).
Apply the double angle formula: \(\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2\cdot\frac{3}{4}}{1-\frac{9}{16}}=\dfrac{\frac{3}{2}}{\frac{7}{16}}=\dfrac{24}{7}\)
\(\Rightarrow2\theta=\tan^{-1}\dfrac{24}{7}\Rightarrow2\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}\)
4. Prove that \(\sin^{-1}\dfrac{8}{17}+\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{77}{36}\).
Solution:
Let \(\alpha=\sin^{-1}\dfrac{8}{17}\) and \(\beta=\sin^{-1}\dfrac{3}{5}\) (both first quadrant).
\(\cos\alpha=\dfrac{15}{17},\ \tan\alpha=\dfrac{8}{15}\); \(\cos\beta=\dfrac{4}{5},\ \tan\beta=\dfrac{3}{4}\).
\(\tan(\alpha+\beta)=\dfrac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\cdot\frac{3}{4}}=\dfrac{\frac{32+45}{60}}{\frac{60-24}{60}}=\dfrac{77}{36}\Rightarrow\alpha+\beta=\tan^{-1}\dfrac{77}{36}\)
5. Prove that \(\cos^{-1}\dfrac{4}{5}+\cos^{-1}\dfrac{12}{13}=\cos^{-1}\dfrac{33}{65}\).
Solution:
Let \(\alpha=\cos^{-1}\dfrac{4}{5}\) and \(\beta=\cos^{-1}\dfrac{12}{13}\). Then \(\sin\alpha=\dfrac{3}{5}\) and \(\sin\beta=\dfrac{5}{13}\).
Apply the cosine addition formula:
\(\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\dfrac{4}{5}\cdot\dfrac{12}{13}-\dfrac{3}{5}\cdot\dfrac{5}{13}=\dfrac{48-15}{65}=\dfrac{33}{65}\)
\(\Rightarrow\alpha+\beta=\cos^{-1}\dfrac{33}{65}\)
6. Prove that \(\cos^{-1}\dfrac{12}{13}+\sin^{-1}\dfrac{3}{5}=\sin^{-1}\dfrac{56}{65}\).
Solution:
Let \(\alpha=\cos^{-1}\dfrac{12}{13}\) and \(\beta=\sin^{-1}\dfrac{3}{5}\). Then \(\sin\alpha=\dfrac{5}{13}\), \(\cos\beta=\dfrac{4}{5}\).
\(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\dfrac{5}{13}\cdot\dfrac{4}{5}+\dfrac{12}{13}\cdot\dfrac{3}{5}=\dfrac{20+36}{65}=\dfrac{56}{65}\)
\(\Rightarrow\alpha+\beta=\sin^{-1}\dfrac{56}{65}\)
7. Prove that \(\tan^{-1}\dfrac{63}{16}=\sin^{-1}\dfrac{5}{13}+\cos^{-1}\dfrac{3}{5}\).
Solution:
Let \(x=\sin^{-1}\dfrac{5}{13}\) and \(y=\cos^{-1}\dfrac{3}{5}\) (both first quadrant).
\(\cos x=\dfrac{12}{13},\ \tan x=\dfrac{5}{12}\); \(\sin y=\dfrac{4}{5},\ \tan y=\dfrac{4}{3}\).
\(\tan(x+y)=\dfrac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\cdot\frac{4}{3}}=\dfrac{\frac{5+16}{12}}{\frac{36-20}{36}}=\dfrac{\frac{21}{12}}{\frac{16}{36}}=\dfrac{7}{4}\cdot\dfrac{9}{4}=\dfrac{63}{16}\Rightarrow x+y=\tan^{-1}\dfrac{63}{16}\)
8. Prove that \(\tan^{-1}\sqrt{x}=\dfrac{1}{2}\cos^{-1}\!\left(\dfrac{1-x}{1+x}\right),\ x\in[0,1]\).
Solution:
Let \(\tan^{-1}\sqrt{x}=\theta\), so \(\sqrt{x}=\tan\theta\) and \(x=\tan^2\theta\).
R.H.S. \(=\dfrac{1}{2}\cos^{-1}\!\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1}{2}\cos^{-1}(\cos2\theta)=\dfrac{1}{2}\cdot2\theta=\theta=\tan^{-1}\sqrt{x}=\) L.H.S.
9. Prove that \(\cot^{-1}\!\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\dfrac{x}{2},\ x\in\left(0,\dfrac{\pi}{4}\right)\).
Solution:
Express \(1+\sin x=\left(\cos\dfrac{x}{2}+\sin\dfrac{x}{2}\right)^2\) and \(1-\sin x=\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)^2\). Substituting and simplifying the square roots:
\(\cot^{-1}\!\left(\dfrac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right)=\cot^{-1}\!\left(\cot\dfrac{x}{2}\right)=\dfrac{x}{2}\)
10. Prove that \(\tan^{-1}\!\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x,\ -\dfrac{1}{\sqrt{2}}\leq x\leq1\).
Solution:
Put \(x=\cos2\theta\) (so \(\theta=\dfrac{1}{2}\cos^{-1}x\)). Use \(1+\cos2\theta=2\cos^2\theta\) and \(1-\cos2\theta=2\sin^2\theta\):
L.H.S. \(=\tan^{-1}\!\left(\dfrac{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}\right)=\tan^{-1}\!\left(\dfrac{1-\tan\theta}{1+\tan\theta}\right)=\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{\pi}{4}-\theta=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x\)
11. Solve: \(2\tan^{-1}(\cos x)=\tan^{-1}(2\operatorname{cosec}x)\).
Solution:
Apply \(2\tan^{-1}t=\tan^{-1}\dfrac{2t}{1-t^2}\) to the left side:
\(\tan^{-1}\!\left(\dfrac{2\cos x}{1-\cos^2x}\right)=\tan^{-1}\!\left(\dfrac{2}{\sin x}\right)\Rightarrow\dfrac{2\cos x}{\sin^2x}=\dfrac{2}{\sin x}\)
Dividing both sides by \(\dfrac{2}{\sin x}\): \(\dfrac{\cos x}{\sin x}=1\Rightarrow\cot x=1=\cot\dfrac{\pi}{4}\Rightarrow x=\dfrac{\pi}{4}\)
12. Solve: \(\tan^{-1}\!\left(\dfrac{1-x}{1+x}\right)=\dfrac{1}{2}\tan^{-1}x,\ (x>0)\).
Solution:
Substitute \(x=\tan\theta\). The equation becomes:
\(\tan^{-1}\!\left(\dfrac{1-\tan\theta}{1+\tan\theta}\right)=\dfrac{\theta}{2}\Rightarrow\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{\theta}{2}\Rightarrow\dfrac{\pi}{4}-\theta=\dfrac{\theta}{2}\)
\(\Rightarrow\dfrac{3\theta}{2}=\dfrac{\pi}{4}\Rightarrow\theta=\dfrac{\pi}{6}\Rightarrow x=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}\)
13. \(\sin(\tan^{-1}x),\ |x|<1\) is equal to: (A) \(\dfrac{x}{\sqrt{1-x^2}}\) (B) \(\dfrac{1}{\sqrt{1-x^2}}\) (C) \(\dfrac{1}{\sqrt{1+x^2}}\) (D) \(\dfrac{x}{\sqrt{1+x^2}}\)
Solution:
Let \(\theta=\tan^{-1}x\) so \(x=\tan\theta\). Then:
\(\sin\theta=\dfrac{1}{\operatorname{cosec}\theta}=\dfrac{1}{\sqrt{1+\cot^2\theta}}=\dfrac{1}{\sqrt{1+\frac{1}{\tan^2\theta}}}=\dfrac{1}{\sqrt{\frac{x^2+1}{x^2}}}=\dfrac{x}{\sqrt{x^2+1}}\)
Correct answer: (D)
14. \(\sin^{-1}(1-x)-2\sin^{-1}x=\dfrac{\pi}{2}\), then \(x\) equals: (A) \(0,\dfrac{1}{2}\) (B) \(1,\dfrac{1}{2}\) (C) \(0\) (D) \(\dfrac{1}{2}\)
Solution:
Let \(\sin^{-1}x=\theta\) (so \(x=\sin\theta\)). The equation becomes:
\(\sin^{-1}(1-x)=\dfrac{\pi}{2}+2\theta\Rightarrow1-x=\sin\!\left(\dfrac{\pi}{2}+2\theta\right)=\cos2\theta=1-2\sin^2\theta=1-2x^2\)
\(\Rightarrow x=2x^2\Rightarrow x(2x-1)=0\Rightarrow x=0\) or \(x=\dfrac{1}{2}\).
Testing \(x=\dfrac{1}{2}\): \(\sin^{-1}\dfrac{1}{2}-2\sin^{-1}\dfrac{1}{2}=-\dfrac{\pi}{6}\neq\dfrac{\pi}{2}\). Rejected.
Testing \(x=0\): \(\sin^{-1}1-0=\dfrac{\pi}{2}\). ✓
Correct answer: (C)
Solution:
Divide numerator and denominator by \(\cos x\), then apply the tangent subtraction formula:
\(=\tan^{-1}\!\left(\dfrac{1-\tan x}{1+\tan x}\right)=\tan^{-1}\!\left(\dfrac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\right)=\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-x\right)=\dfrac{\pi}{4}-x\)
Divide numerator and denominator by \(\cos x\), then apply the tangent subtraction formula:
\(=\tan^{-1}\!\left(\dfrac{1-\tan x}{1+\tan x}\right)=\tan^{-1}\!\left(\dfrac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\right)=\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-x\right)=\dfrac{\pi}{4}-x\)
6. Write \(\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}},\ |x|
Solution:
Substitute \(x=a\sin\theta\) (so \(\theta=\sin^{-1}\dfrac{x}{a}\)):
\(\tan^{-1}\!\left(\dfrac{a\sin\theta}{\sqrt{a^2-a^2\sin^2\theta}}\right)=\tan^{-1}\!\left(\dfrac{a\sin\theta}{a\cos\theta}\right)=\tan^{-1}(\tan\theta)=\theta=\sin^{-1}\dfrac{x}{a}\)
7. Write \(\tan^{-1}\!\left(\dfrac{3a^2x-x^3}{a^3-3ax^2}\right),\ a>0,\ -\dfrac{a}{\sqrt{3}}\leq x\leq\dfrac{a}{\sqrt{3}}\) in simplest form.
Solution:
Divide numerator and denominator by \(a^3\), then substitute \(\dfrac{x}{a}=\tan\theta\):
\(\tan^{-1}\!\left(\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\right)=\tan^{-1}(\tan3\theta)=3\theta=3\tan^{-1}\dfrac{x}{a}\)
8. Find the value of \(\tan^{-1}\!\left[2\cos\!\left(2\sin^{-1}\dfrac{1}{2}\right)\right]\).
Solution:
Work from the inside outward, substituting known values step by step:
\(=\tan^{-1}\!\left[2\cos\!\left(2\cdot\dfrac{\pi}{6}\right)\right]=\tan^{-1}\!\left[2\cos\dfrac{\pi}{3}\right]=\tan^{-1}\!\left[2\cdot\dfrac{1}{2}\right]=\tan^{-1}1=\dfrac{\pi}{4}\)
9. Find the value of \(\tan\dfrac{1}{2}\!\left[\sin^{-1}\dfrac{2x}{1+x^2}+\cos^{-1}\dfrac{1-y^2}{1+y^2}\right]\), \(|x|<1,y>0,xy<1\).
Solution:
Substitute \(x=\tan\theta\) and \(y=\tan\phi\). Use the double-angle identities \(\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}\) and \(\cos2\phi=\dfrac{1-\tan^2\phi}{1+\tan^2\phi}\):
\(=\tan\!\left[\dfrac{1}{2}\sin^{-1}(\sin2\theta)+\dfrac{1}{2}\cos^{-1}(\cos2\phi)\right]=\tan(\theta+\phi)=\dfrac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}=\dfrac{x+y}{1-xy}\)
10. Find the value of \(\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)\).
Solution:
Since \(\dfrac{2\pi}{3}\notin\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\), we cannot apply the identity directly. Rewrite using the supplementary angle:
\(\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)=\sin^{-1}\!\left(\sin\!\left(\pi-\dfrac{\pi}{3}\right)\right)=\sin^{-1}\!\left(\sin\dfrac{\pi}{3}\right)=\dfrac{\pi}{3}\)
11. Find the value of \(\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)\).
Solution:
Since \(\dfrac{3\pi}{4}\notin\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), rewrite using the identity \(\tan(\pi-\theta)=-\tan\theta\):
\(\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)=\tan^{-1}\!\left(\tan\!\left(\pi-\dfrac{\pi}{4}\right)\right)=\tan^{-1}\!\left(-\tan\dfrac{\pi}{4}\right)=-\tan^{-1}\tan\dfrac{\pi}{4}=-\dfrac{\pi}{4}\)
12. Find the value of \(\tan\!\left(\sin^{-1}\dfrac{3}{5}+\cot^{-1}\dfrac{3}{2}\right)\).
Solution:
Let \(\sin^{-1}\dfrac{3}{5}=x\) and \(\cot^{-1}\dfrac{3}{2}=y\). Both angles are in the first quadrant.
From \(\sin x=\dfrac{3}{5}\): \(\cos x=\dfrac{4}{5}\), so \(\tan x=\dfrac{3}{4}\).
From \(\cot y=\dfrac{3}{2}\): \(\tan y=\dfrac{2}{3}\).
\(\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\dfrac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot\frac{2}{3}}=\dfrac{\frac{17}{12}}{\frac{1}{2}}=\dfrac{17}{6}\)
13. \(\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)\) is equal to: (A) \(\dfrac{7\pi}{6}\) (B) \(\dfrac{5\pi}{6}\) (C) \(\dfrac{\pi}{3}\) (D) \(\dfrac{\pi}{6}\)
Solution:
Since \(\cos\dfrac{7\pi}{6}<0\), the result must lie in \(\left[\dfrac{\pi}{2},\pi\right]\). Use \(\cos(2\pi-\theta)=\cos\theta\):
\(\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)=\cos^{-1}\!\left(\cos\!\left(2\pi-\dfrac{7\pi}{6}\right)\right)=\dfrac{5\pi}{6}\)
Correct answer: (B)
14. \(\sin\!\left(\dfrac{\pi}{3}-\sin^{-1}\!\left(-\dfrac{1}{2}\right)\right)\) is equal to: (A) \(\dfrac{1}{2}\) (B) \(\dfrac{1}{3}\) (C) \(\dfrac{1}{4}\) (D) 1
Solution:
Use \(\sin^{-1}(-x)=-\sin^{-1}x\) to simplify the inner expression:
\(\sin\!\left(\dfrac{\pi}{3}+\sin^{-1}\dfrac{1}{2}\right)=\sin\!\left(\dfrac{\pi}{3}+\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{2}=1\)
Correct answer: (D)
15. \(\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})\) is equal to: (A) \(\pi\) (B) \(-\dfrac{\pi}{2}\) (C) 0 (D) \(2\sqrt{3}\)
Solution:
Use \(\cot^{-1}(-x)=\pi-\cot^{-1}x\):
\(\tan^{-1}\sqrt{3}-(\pi-\cot^{-1}\sqrt{3})=\dfrac{\pi}{3}-\pi+\dfrac{\pi}{6}=\dfrac{2\pi-6\pi+\pi}{6}=-\dfrac{3\pi}{6}=-\dfrac{\pi}{2}\)
Correct answer: (B)
Solution:
Substitute \(x=a\sin\theta\) (so \(\theta=\sin^{-1}\dfrac{x}{a}\)):
\(\tan^{-1}\!\left(\dfrac{a\sin\theta}{\sqrt{a^2-a^2\sin^2\theta}}\right)=\tan^{-1}\!\left(\dfrac{a\sin\theta}{a\cos\theta}\right)=\tan^{-1}(\tan\theta)=\theta=\sin^{-1}\dfrac{x}{a}\)
Substitute \(x=a\sin\theta\) (so \(\theta=\sin^{-1}\dfrac{x}{a}\)):
\(\tan^{-1}\!\left(\dfrac{a\sin\theta}{\sqrt{a^2-a^2\sin^2\theta}}\right)=\tan^{-1}\!\left(\dfrac{a\sin\theta}{a\cos\theta}\right)=\tan^{-1}(\tan\theta)=\theta=\sin^{-1}\dfrac{x}{a}\)
7. Write \(\tan^{-1}\!\left(\dfrac{3a^2x-x^3}{a^3-3ax^2}\right),\ a>0,\ -\dfrac{a}{\sqrt{3}}\leq x\leq\dfrac{a}{\sqrt{3}}\) in simplest form.
Solution:
Divide numerator and denominator by \(a^3\), then substitute \(\dfrac{x}{a}=\tan\theta\):
\(\tan^{-1}\!\left(\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\right)=\tan^{-1}(\tan3\theta)=3\theta=3\tan^{-1}\dfrac{x}{a}\)
Divide numerator and denominator by \(a^3\), then substitute \(\dfrac{x}{a}=\tan\theta\):
\(\tan^{-1}\!\left(\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\right)=\tan^{-1}(\tan3\theta)=3\theta=3\tan^{-1}\dfrac{x}{a}\)
8. Find the value of \(\tan^{-1}\!\left[2\cos\!\left(2\sin^{-1}\dfrac{1}{2}\right)\right]\).
Solution:
Work from the inside outward, substituting known values step by step:
\(=\tan^{-1}\!\left[2\cos\!\left(2\cdot\dfrac{\pi}{6}\right)\right]=\tan^{-1}\!\left[2\cos\dfrac{\pi}{3}\right]=\tan^{-1}\!\left[2\cdot\dfrac{1}{2}\right]=\tan^{-1}1=\dfrac{\pi}{4}\)
Work from the inside outward, substituting known values step by step:
\(=\tan^{-1}\!\left[2\cos\!\left(2\cdot\dfrac{\pi}{6}\right)\right]=\tan^{-1}\!\left[2\cos\dfrac{\pi}{3}\right]=\tan^{-1}\!\left[2\cdot\dfrac{1}{2}\right]=\tan^{-1}1=\dfrac{\pi}{4}\)
9. Find the value of \(\tan\dfrac{1}{2}\!\left[\sin^{-1}\dfrac{2x}{1+x^2}+\cos^{-1}\dfrac{1-y^2}{1+y^2}\right]\), \(|x|<1,y>0,xy<1\).
Solution:
Substitute \(x=\tan\theta\) and \(y=\tan\phi\). Use the double-angle identities \(\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}\) and \(\cos2\phi=\dfrac{1-\tan^2\phi}{1+\tan^2\phi}\):
\(=\tan\!\left[\dfrac{1}{2}\sin^{-1}(\sin2\theta)+\dfrac{1}{2}\cos^{-1}(\cos2\phi)\right]=\tan(\theta+\phi)=\dfrac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}=\dfrac{x+y}{1-xy}\)
Substitute \(x=\tan\theta\) and \(y=\tan\phi\). Use the double-angle identities \(\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}\) and \(\cos2\phi=\dfrac{1-\tan^2\phi}{1+\tan^2\phi}\):
\(=\tan\!\left[\dfrac{1}{2}\sin^{-1}(\sin2\theta)+\dfrac{1}{2}\cos^{-1}(\cos2\phi)\right]=\tan(\theta+\phi)=\dfrac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}=\dfrac{x+y}{1-xy}\)
10. Find the value of \(\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)\).
Solution:
Since \(\dfrac{2\pi}{3}\notin\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\), we cannot apply the identity directly. Rewrite using the supplementary angle:
\(\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)=\sin^{-1}\!\left(\sin\!\left(\pi-\dfrac{\pi}{3}\right)\right)=\sin^{-1}\!\left(\sin\dfrac{\pi}{3}\right)=\dfrac{\pi}{3}\)
Since \(\dfrac{2\pi}{3}\notin\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\), we cannot apply the identity directly. Rewrite using the supplementary angle:
\(\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)=\sin^{-1}\!\left(\sin\!\left(\pi-\dfrac{\pi}{3}\right)\right)=\sin^{-1}\!\left(\sin\dfrac{\pi}{3}\right)=\dfrac{\pi}{3}\)
11. Find the value of \(\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)\).
Solution:
Since \(\dfrac{3\pi}{4}\notin\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), rewrite using the identity \(\tan(\pi-\theta)=-\tan\theta\):
\(\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)=\tan^{-1}\!\left(\tan\!\left(\pi-\dfrac{\pi}{4}\right)\right)=\tan^{-1}\!\left(-\tan\dfrac{\pi}{4}\right)=-\tan^{-1}\tan\dfrac{\pi}{4}=-\dfrac{\pi}{4}\)
Since \(\dfrac{3\pi}{4}\notin\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), rewrite using the identity \(\tan(\pi-\theta)=-\tan\theta\):
\(\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)=\tan^{-1}\!\left(\tan\!\left(\pi-\dfrac{\pi}{4}\right)\right)=\tan^{-1}\!\left(-\tan\dfrac{\pi}{4}\right)=-\tan^{-1}\tan\dfrac{\pi}{4}=-\dfrac{\pi}{4}\)
12. Find the value of \(\tan\!\left(\sin^{-1}\dfrac{3}{5}+\cot^{-1}\dfrac{3}{2}\right)\).
Solution:
Let \(\sin^{-1}\dfrac{3}{5}=x\) and \(\cot^{-1}\dfrac{3}{2}=y\). Both angles are in the first quadrant.
From \(\sin x=\dfrac{3}{5}\): \(\cos x=\dfrac{4}{5}\), so \(\tan x=\dfrac{3}{4}\).
From \(\cot y=\dfrac{3}{2}\): \(\tan y=\dfrac{2}{3}\).
\(\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\dfrac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot\frac{2}{3}}=\dfrac{\frac{17}{12}}{\frac{1}{2}}=\dfrac{17}{6}\)
Let \(\sin^{-1}\dfrac{3}{5}=x\) and \(\cot^{-1}\dfrac{3}{2}=y\). Both angles are in the first quadrant.
From \(\sin x=\dfrac{3}{5}\): \(\cos x=\dfrac{4}{5}\), so \(\tan x=\dfrac{3}{4}\).
From \(\cot y=\dfrac{3}{2}\): \(\tan y=\dfrac{2}{3}\).
\(\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\dfrac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot\frac{2}{3}}=\dfrac{\frac{17}{12}}{\frac{1}{2}}=\dfrac{17}{6}\)
13. \(\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)\) is equal to: (A) \(\dfrac{7\pi}{6}\) (B) \(\dfrac{5\pi}{6}\) (C) \(\dfrac{\pi}{3}\) (D) \(\dfrac{\pi}{6}\)
Solution:
Since \(\cos\dfrac{7\pi}{6}<0\), the result must lie in \(\left[\dfrac{\pi}{2},\pi\right]\). Use \(\cos(2\pi-\theta)=\cos\theta\):
\(\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)=\cos^{-1}\!\left(\cos\!\left(2\pi-\dfrac{7\pi}{6}\right)\right)=\dfrac{5\pi}{6}\)
Correct answer: (B)
Since \(\cos\dfrac{7\pi}{6}<0\), the result must lie in \(\left[\dfrac{\pi}{2},\pi\right]\). Use \(\cos(2\pi-\theta)=\cos\theta\):
\(\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)=\cos^{-1}\!\left(\cos\!\left(2\pi-\dfrac{7\pi}{6}\right)\right)=\dfrac{5\pi}{6}\)
Correct answer: (B)
14. \(\sin\!\left(\dfrac{\pi}{3}-\sin^{-1}\!\left(-\dfrac{1}{2}\right)\right)\) is equal to: (A) \(\dfrac{1}{2}\) (B) \(\dfrac{1}{3}\) (C) \(\dfrac{1}{4}\) (D) 1
Solution:
Use \(\sin^{-1}(-x)=-\sin^{-1}x\) to simplify the inner expression:
\(\sin\!\left(\dfrac{\pi}{3}+\sin^{-1}\dfrac{1}{2}\right)=\sin\!\left(\dfrac{\pi}{3}+\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{2}=1\)
Correct answer: (D)
Use \(\sin^{-1}(-x)=-\sin^{-1}x\) to simplify the inner expression:
\(\sin\!\left(\dfrac{\pi}{3}+\sin^{-1}\dfrac{1}{2}\right)=\sin\!\left(\dfrac{\pi}{3}+\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{2}=1\)
Correct answer: (D)
15. \(\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})\) is equal to: (A) \(\pi\) (B) \(-\dfrac{\pi}{2}\) (C) 0 (D) \(2\sqrt{3}\)
Solution:
Use \(\cot^{-1}(-x)=\pi-\cot^{-1}x\):
\(\tan^{-1}\sqrt{3}-(\pi-\cot^{-1}\sqrt{3})=\dfrac{\pi}{3}-\pi+\dfrac{\pi}{6}=\dfrac{2\pi-6\pi+\pi}{6}=-\dfrac{3\pi}{6}=-\dfrac{\pi}{2}\)
Correct answer: (B)
Use \(\cot^{-1}(-x)=\pi-\cot^{-1}x\):
\(\tan^{-1}\sqrt{3}-(\pi-\cot^{-1}\sqrt{3})=\dfrac{\pi}{3}-\pi+\dfrac{\pi}{6}=\dfrac{2\pi-6\pi+\pi}{6}=-\dfrac{3\pi}{6}=-\dfrac{\pi}{2}\)
Correct answer: (B)
Miscellaneous Exercise (Chapter 2)
1. Find the value of \(\cos^{-1}\!\left(\cos\dfrac{13\pi}{6}\right)\).
Solution:
Reduce the angle: \(\cos\dfrac{13\pi}{6}=\cos\!\left(2\pi+\dfrac{\pi}{6}\right)=\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}>0\). The result lies in the first quadrant.
\(\cos^{-1}\!\left(\cos\dfrac{13\pi}{6}\right)=\cos^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{6}\)
Reduce the angle: \(\cos\dfrac{13\pi}{6}=\cos\!\left(2\pi+\dfrac{\pi}{6}\right)=\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}>0\). The result lies in the first quadrant.
\(\cos^{-1}\!\left(\cos\dfrac{13\pi}{6}\right)=\cos^{-1}\dfrac{\sqrt{3}}{2}=\dfrac{\pi}{6}\)
2. Find the value of \(\tan^{-1}\!\left(\tan\dfrac{7\pi}{6}\right)\).
Solution:
Reduce: \(\tan\dfrac{7\pi}{6}=\tan\!\left(\pi+\dfrac{\pi}{6}\right)=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}>0\). The result lies in the first quadrant.
\(\tan^{-1}\!\left(\tan\dfrac{7\pi}{6}\right)=\tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}\)
Reduce: \(\tan\dfrac{7\pi}{6}=\tan\!\left(\pi+\dfrac{\pi}{6}\right)=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}>0\). The result lies in the first quadrant.
\(\tan^{-1}\!\left(\tan\dfrac{7\pi}{6}\right)=\tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}\)
3. Prove that \(2\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}\).
Solution:
Let \(\sin^{-1}\dfrac{3}{5}=\theta\) (first quadrant). Then \(\cos\theta=\dfrac{4}{5}\) and \(\tan\theta=\dfrac{3}{4}\).
Apply the double angle formula: \(\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2\cdot\frac{3}{4}}{1-\frac{9}{16}}=\dfrac{\frac{3}{2}}{\frac{7}{16}}=\dfrac{24}{7}\)
\(\Rightarrow2\theta=\tan^{-1}\dfrac{24}{7}\Rightarrow2\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}\)
Let \(\sin^{-1}\dfrac{3}{5}=\theta\) (first quadrant). Then \(\cos\theta=\dfrac{4}{5}\) and \(\tan\theta=\dfrac{3}{4}\).
Apply the double angle formula: \(\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2\cdot\frac{3}{4}}{1-\frac{9}{16}}=\dfrac{\frac{3}{2}}{\frac{7}{16}}=\dfrac{24}{7}\)
\(\Rightarrow2\theta=\tan^{-1}\dfrac{24}{7}\Rightarrow2\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{24}{7}\)
4. Prove that \(\sin^{-1}\dfrac{8}{17}+\sin^{-1}\dfrac{3}{5}=\tan^{-1}\dfrac{77}{36}\).
Solution:
Let \(\alpha=\sin^{-1}\dfrac{8}{17}\) and \(\beta=\sin^{-1}\dfrac{3}{5}\) (both first quadrant).
\(\cos\alpha=\dfrac{15}{17},\ \tan\alpha=\dfrac{8}{15}\); \(\cos\beta=\dfrac{4}{5},\ \tan\beta=\dfrac{3}{4}\).
\(\tan(\alpha+\beta)=\dfrac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\cdot\frac{3}{4}}=\dfrac{\frac{32+45}{60}}{\frac{60-24}{60}}=\dfrac{77}{36}\Rightarrow\alpha+\beta=\tan^{-1}\dfrac{77}{36}\)
Let \(\alpha=\sin^{-1}\dfrac{8}{17}\) and \(\beta=\sin^{-1}\dfrac{3}{5}\) (both first quadrant).
\(\cos\alpha=\dfrac{15}{17},\ \tan\alpha=\dfrac{8}{15}\); \(\cos\beta=\dfrac{4}{5},\ \tan\beta=\dfrac{3}{4}\).
\(\tan(\alpha+\beta)=\dfrac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\cdot\frac{3}{4}}=\dfrac{\frac{32+45}{60}}{\frac{60-24}{60}}=\dfrac{77}{36}\Rightarrow\alpha+\beta=\tan^{-1}\dfrac{77}{36}\)
5. Prove that \(\cos^{-1}\dfrac{4}{5}+\cos^{-1}\dfrac{12}{13}=\cos^{-1}\dfrac{33}{65}\).
Solution:
Let \(\alpha=\cos^{-1}\dfrac{4}{5}\) and \(\beta=\cos^{-1}\dfrac{12}{13}\). Then \(\sin\alpha=\dfrac{3}{5}\) and \(\sin\beta=\dfrac{5}{13}\).
Apply the cosine addition formula:
\(\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\dfrac{4}{5}\cdot\dfrac{12}{13}-\dfrac{3}{5}\cdot\dfrac{5}{13}=\dfrac{48-15}{65}=\dfrac{33}{65}\)
\(\Rightarrow\alpha+\beta=\cos^{-1}\dfrac{33}{65}\)
Let \(\alpha=\cos^{-1}\dfrac{4}{5}\) and \(\beta=\cos^{-1}\dfrac{12}{13}\). Then \(\sin\alpha=\dfrac{3}{5}\) and \(\sin\beta=\dfrac{5}{13}\).
Apply the cosine addition formula:
\(\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\dfrac{4}{5}\cdot\dfrac{12}{13}-\dfrac{3}{5}\cdot\dfrac{5}{13}=\dfrac{48-15}{65}=\dfrac{33}{65}\)
\(\Rightarrow\alpha+\beta=\cos^{-1}\dfrac{33}{65}\)
6. Prove that \(\cos^{-1}\dfrac{12}{13}+\sin^{-1}\dfrac{3}{5}=\sin^{-1}\dfrac{56}{65}\).
Solution:
Let \(\alpha=\cos^{-1}\dfrac{12}{13}\) and \(\beta=\sin^{-1}\dfrac{3}{5}\). Then \(\sin\alpha=\dfrac{5}{13}\), \(\cos\beta=\dfrac{4}{5}\).
\(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\dfrac{5}{13}\cdot\dfrac{4}{5}+\dfrac{12}{13}\cdot\dfrac{3}{5}=\dfrac{20+36}{65}=\dfrac{56}{65}\)
\(\Rightarrow\alpha+\beta=\sin^{-1}\dfrac{56}{65}\)
Let \(\alpha=\cos^{-1}\dfrac{12}{13}\) and \(\beta=\sin^{-1}\dfrac{3}{5}\). Then \(\sin\alpha=\dfrac{5}{13}\), \(\cos\beta=\dfrac{4}{5}\).
\(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\dfrac{5}{13}\cdot\dfrac{4}{5}+\dfrac{12}{13}\cdot\dfrac{3}{5}=\dfrac{20+36}{65}=\dfrac{56}{65}\)
\(\Rightarrow\alpha+\beta=\sin^{-1}\dfrac{56}{65}\)
7. Prove that \(\tan^{-1}\dfrac{63}{16}=\sin^{-1}\dfrac{5}{13}+\cos^{-1}\dfrac{3}{5}\).
Solution:
Let \(x=\sin^{-1}\dfrac{5}{13}\) and \(y=\cos^{-1}\dfrac{3}{5}\) (both first quadrant).
\(\cos x=\dfrac{12}{13},\ \tan x=\dfrac{5}{12}\); \(\sin y=\dfrac{4}{5},\ \tan y=\dfrac{4}{3}\).
\(\tan(x+y)=\dfrac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\cdot\frac{4}{3}}=\dfrac{\frac{5+16}{12}}{\frac{36-20}{36}}=\dfrac{\frac{21}{12}}{\frac{16}{36}}=\dfrac{7}{4}\cdot\dfrac{9}{4}=\dfrac{63}{16}\Rightarrow x+y=\tan^{-1}\dfrac{63}{16}\)
Let \(x=\sin^{-1}\dfrac{5}{13}\) and \(y=\cos^{-1}\dfrac{3}{5}\) (both first quadrant).
\(\cos x=\dfrac{12}{13},\ \tan x=\dfrac{5}{12}\); \(\sin y=\dfrac{4}{5},\ \tan y=\dfrac{4}{3}\).
\(\tan(x+y)=\dfrac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\cdot\frac{4}{3}}=\dfrac{\frac{5+16}{12}}{\frac{36-20}{36}}=\dfrac{\frac{21}{12}}{\frac{16}{36}}=\dfrac{7}{4}\cdot\dfrac{9}{4}=\dfrac{63}{16}\Rightarrow x+y=\tan^{-1}\dfrac{63}{16}\)
8. Prove that \(\tan^{-1}\sqrt{x}=\dfrac{1}{2}\cos^{-1}\!\left(\dfrac{1-x}{1+x}\right),\ x\in[0,1]\).
Solution:
Let \(\tan^{-1}\sqrt{x}=\theta\), so \(\sqrt{x}=\tan\theta\) and \(x=\tan^2\theta\).
R.H.S. \(=\dfrac{1}{2}\cos^{-1}\!\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1}{2}\cos^{-1}(\cos2\theta)=\dfrac{1}{2}\cdot2\theta=\theta=\tan^{-1}\sqrt{x}=\) L.H.S.
Let \(\tan^{-1}\sqrt{x}=\theta\), so \(\sqrt{x}=\tan\theta\) and \(x=\tan^2\theta\).
R.H.S. \(=\dfrac{1}{2}\cos^{-1}\!\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1}{2}\cos^{-1}(\cos2\theta)=\dfrac{1}{2}\cdot2\theta=\theta=\tan^{-1}\sqrt{x}=\) L.H.S.
9. Prove that \(\cot^{-1}\!\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\dfrac{x}{2},\ x\in\left(0,\dfrac{\pi}{4}\right)\).
Solution:
Express \(1+\sin x=\left(\cos\dfrac{x}{2}+\sin\dfrac{x}{2}\right)^2\) and \(1-\sin x=\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)^2\). Substituting and simplifying the square roots:
\(\cot^{-1}\!\left(\dfrac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right)=\cot^{-1}\!\left(\cot\dfrac{x}{2}\right)=\dfrac{x}{2}\)
Express \(1+\sin x=\left(\cos\dfrac{x}{2}+\sin\dfrac{x}{2}\right)^2\) and \(1-\sin x=\left(\cos\dfrac{x}{2}-\sin\dfrac{x}{2}\right)^2\). Substituting and simplifying the square roots:
\(\cot^{-1}\!\left(\dfrac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}\right)=\cot^{-1}\!\left(\cot\dfrac{x}{2}\right)=\dfrac{x}{2}\)
10. Prove that \(\tan^{-1}\!\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x,\ -\dfrac{1}{\sqrt{2}}\leq x\leq1\).
Solution:
Put \(x=\cos2\theta\) (so \(\theta=\dfrac{1}{2}\cos^{-1}x\)). Use \(1+\cos2\theta=2\cos^2\theta\) and \(1-\cos2\theta=2\sin^2\theta\):
L.H.S. \(=\tan^{-1}\!\left(\dfrac{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}\right)=\tan^{-1}\!\left(\dfrac{1-\tan\theta}{1+\tan\theta}\right)=\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{\pi}{4}-\theta=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x\)
Put \(x=\cos2\theta\) (so \(\theta=\dfrac{1}{2}\cos^{-1}x\)). Use \(1+\cos2\theta=2\cos^2\theta\) and \(1-\cos2\theta=2\sin^2\theta\):
L.H.S. \(=\tan^{-1}\!\left(\dfrac{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}\right)=\tan^{-1}\!\left(\dfrac{1-\tan\theta}{1+\tan\theta}\right)=\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{\pi}{4}-\theta=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x\)
11. Solve: \(2\tan^{-1}(\cos x)=\tan^{-1}(2\operatorname{cosec}x)\).
Solution:
Apply \(2\tan^{-1}t=\tan^{-1}\dfrac{2t}{1-t^2}\) to the left side:
\(\tan^{-1}\!\left(\dfrac{2\cos x}{1-\cos^2x}\right)=\tan^{-1}\!\left(\dfrac{2}{\sin x}\right)\Rightarrow\dfrac{2\cos x}{\sin^2x}=\dfrac{2}{\sin x}\)
Dividing both sides by \(\dfrac{2}{\sin x}\): \(\dfrac{\cos x}{\sin x}=1\Rightarrow\cot x=1=\cot\dfrac{\pi}{4}\Rightarrow x=\dfrac{\pi}{4}\)
Apply \(2\tan^{-1}t=\tan^{-1}\dfrac{2t}{1-t^2}\) to the left side:
\(\tan^{-1}\!\left(\dfrac{2\cos x}{1-\cos^2x}\right)=\tan^{-1}\!\left(\dfrac{2}{\sin x}\right)\Rightarrow\dfrac{2\cos x}{\sin^2x}=\dfrac{2}{\sin x}\)
Dividing both sides by \(\dfrac{2}{\sin x}\): \(\dfrac{\cos x}{\sin x}=1\Rightarrow\cot x=1=\cot\dfrac{\pi}{4}\Rightarrow x=\dfrac{\pi}{4}\)
12. Solve: \(\tan^{-1}\!\left(\dfrac{1-x}{1+x}\right)=\dfrac{1}{2}\tan^{-1}x,\ (x>0)\).
Solution:
Substitute \(x=\tan\theta\). The equation becomes:
\(\tan^{-1}\!\left(\dfrac{1-\tan\theta}{1+\tan\theta}\right)=\dfrac{\theta}{2}\Rightarrow\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{\theta}{2}\Rightarrow\dfrac{\pi}{4}-\theta=\dfrac{\theta}{2}\)
\(\Rightarrow\dfrac{3\theta}{2}=\dfrac{\pi}{4}\Rightarrow\theta=\dfrac{\pi}{6}\Rightarrow x=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}\)
Substitute \(x=\tan\theta\). The equation becomes:
\(\tan^{-1}\!\left(\dfrac{1-\tan\theta}{1+\tan\theta}\right)=\dfrac{\theta}{2}\Rightarrow\tan^{-1}\tan\!\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{\theta}{2}\Rightarrow\dfrac{\pi}{4}-\theta=\dfrac{\theta}{2}\)
\(\Rightarrow\dfrac{3\theta}{2}=\dfrac{\pi}{4}\Rightarrow\theta=\dfrac{\pi}{6}\Rightarrow x=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt{3}}\)
13. \(\sin(\tan^{-1}x),\ |x|<1\) is equal to: (A) \(\dfrac{x}{\sqrt{1-x^2}}\) (B) \(\dfrac{1}{\sqrt{1-x^2}}\) (C) \(\dfrac{1}{\sqrt{1+x^2}}\) (D) \(\dfrac{x}{\sqrt{1+x^2}}\)
Solution:
Let \(\theta=\tan^{-1}x\) so \(x=\tan\theta\). Then:
\(\sin\theta=\dfrac{1}{\operatorname{cosec}\theta}=\dfrac{1}{\sqrt{1+\cot^2\theta}}=\dfrac{1}{\sqrt{1+\frac{1}{\tan^2\theta}}}=\dfrac{1}{\sqrt{\frac{x^2+1}{x^2}}}=\dfrac{x}{\sqrt{x^2+1}}\)
Correct answer: (D)
Let \(\theta=\tan^{-1}x\) so \(x=\tan\theta\). Then:
\(\sin\theta=\dfrac{1}{\operatorname{cosec}\theta}=\dfrac{1}{\sqrt{1+\cot^2\theta}}=\dfrac{1}{\sqrt{1+\frac{1}{\tan^2\theta}}}=\dfrac{1}{\sqrt{\frac{x^2+1}{x^2}}}=\dfrac{x}{\sqrt{x^2+1}}\)
Correct answer: (D)
14. \(\sin^{-1}(1-x)-2\sin^{-1}x=\dfrac{\pi}{2}\), then \(x\) equals: (A) \(0,\dfrac{1}{2}\) (B) \(1,\dfrac{1}{2}\) (C) \(0\) (D) \(\dfrac{1}{2}\)
Solution:
Let \(\sin^{-1}x=\theta\) (so \(x=\sin\theta\)). The equation becomes:
\(\sin^{-1}(1-x)=\dfrac{\pi}{2}+2\theta\Rightarrow1-x=\sin\!\left(\dfrac{\pi}{2}+2\theta\right)=\cos2\theta=1-2\sin^2\theta=1-2x^2\)
\(\Rightarrow x=2x^2\Rightarrow x(2x-1)=0\Rightarrow x=0\) or \(x=\dfrac{1}{2}\).
Testing \(x=\dfrac{1}{2}\): \(\sin^{-1}\dfrac{1}{2}-2\sin^{-1}\dfrac{1}{2}=-\dfrac{\pi}{6}\neq\dfrac{\pi}{2}\). Rejected.
Testing \(x=0\): \(\sin^{-1}1-0=\dfrac{\pi}{2}\). ✓
Correct answer: (C)
Let \(\sin^{-1}x=\theta\) (so \(x=\sin\theta\)). The equation becomes:
\(\sin^{-1}(1-x)=\dfrac{\pi}{2}+2\theta\Rightarrow1-x=\sin\!\left(\dfrac{\pi}{2}+2\theta\right)=\cos2\theta=1-2\sin^2\theta=1-2x^2\)
\(\Rightarrow x=2x^2\Rightarrow x(2x-1)=0\Rightarrow x=0\) or \(x=\dfrac{1}{2}\).
Testing \(x=\dfrac{1}{2}\): \(\sin^{-1}\dfrac{1}{2}-2\sin^{-1}\dfrac{1}{2}=-\dfrac{\pi}{6}\neq\dfrac{\pi}{2}\). Rejected.
Testing \(x=0\): \(\sin^{-1}1-0=\dfrac{\pi}{2}\). ✓
Correct answer: (C)
