Ncert Solutions Class 12 Chapter 2 Inverse Trigonometric Functions Exercise M Question 12

Solve the following equations:

12. (tan^{-1}{frac{1-x}{1+x}})(=frac{1}{2}tan^{-1}x,;(x>0))


Inverse Trigonometric Functions

Miscellaneous Exercise

Ncert solutions class 12 chapter 2 Miscellaneous exercise

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Class 12

Inverse Trigonometric Functions

Miscellaneous Exercise

Find the value of the following:

1. (cos^{-1}(cosfrac{13pi}{6})).

2. (tan^{-1}(tanfrac{7pi}{6})).

3.  (2sin^{-1}{frac{3}{5}})(=tan^{-1}{frac{24}{7}}).

4. (sin^{-1}{frac{8}{17}}+sin^{-1}{frac{3}{5}})(=tan^{-1}{frac{77}{36}}).

5. (cos^{-1}{frac{4}{5}}+cos^{-1}{frac{12}{13}})(=cos^{-1}{frac{33}{65}}).

6. (cos^{-1}{frac{12}{13}}+sin^{-1}{frac{3}{5}})(=sin^{-1}{frac{56}{65}}).

7. (tan^{-1}{frac{63}{16}}=sin^{-1}{frac{5}{13}})(+cos^{-1}{frac{3}{5}}).

Prove that:

8. (tan^{-1}{sqrt{x}}=frac{1}{2}cos^{-1}{frac{1-x}{1+x}}).

9. (cot^{-1}(frac{sqrt{1+sinx}+sqrt{1-sinx}}{sqrt{1+sinx}+sqrt{1-sinx}}))(=frac{x}{2},;xin(0,frac{pi}{4})).

10. (tan^{-1}(frac{sqrt{1+x}+sqrt{1-x}}{sqrt{1+x}+sqrt{1-x}}))(=frac{pi}{4}-frac{1}{2}cos^{-1}x),(;-frac{1}{sqrt2}le{x}le{1}).

Solve the following equations:

11. (2tan^{-1}(cosx)=tan^{-1}(2cosecx)).

12. (tan^{-1}{frac{1-x}{1+x}})(=frac{1}{2}tan^{-1}x,;(x>0))

13. (sin(tan^{-1}x),|x|<1) is equal to:

(A) (frac{x}{sqrt{1-x^2}})

(B) (frac{1}{sqrt{1-x^2}})

(C) (frac{1}{sqrt{1+x^2}})

(D) (frac{x}{sqrt{1+x^2}})

14. (sin^{-1}(1-x)-2sin^{-1}x)(=frac{pi}{2}), then x is equal to:

(A) (0,frac{1}{2})

(B) (1,frac{1}{2})

(C) (0)

(D) (frac{1}{2})



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