Dr.Harswaroop Sharma Mathematics Solution Class 12 Chapter 2 प्रतिलोम त्रिकोणमितीय फलन (Inverse Trigonometric Function) Exercise 2B

   

प्रश्नावली - 2B



Question 1

सिद्ध कीजिए कि :
1. $\tan ^{-1} \frac{2}{11}+\cot ^{-1} \frac{24}{7}=\tan ^{-1} \frac{1}{2}$.







Question 2

2. $\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}=\cot ^{-1} \frac{9}{2}$.






Question 3

3. $\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$.








Question 4

4. $\sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{63}{65}=2 \tan ^{-1} \frac{1}{5}$.






Question 5

5. $\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{63}{65}$.








Question 6

6. $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq 1$






Question 7

7. $\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{56}{65}$.







Question 8

8. $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \cos ^{-1} \frac{3}{5}$.






Question 9

9. $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$.






Question 10

10. $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$.






Question 11

11. $\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{5}=\tan ^{-1} \frac{27}{11}$.







Question 12

12. (i) $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2} ; x \in\left(0, \frac{\pi}{4}\right)$
(ii) $\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}=\tan ^{-1} \frac{63}{16}$.






Question 13

13. $\cos ^{-1} x=2 \sin ^{-1} \sqrt{\frac{1-x}{2}}=2 \cos ^{-1} \sqrt{\frac{1+x}{2}}$.







Question 14

14. $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$






Question 15

15. $\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right), x \in[0,1]$









Question 16

16. यदि $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$, तो सिद्ध कीजिए :
(i) $x \sqrt{1-y^2}+y \sqrt{1-x^2}=1$
(ii) $\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^2}=\frac{\pi}{2}$




Question 17

17. यदि $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\pi$, तो सिद्ध कीजिए कि :
$x \sqrt{1-x^2}+y \sqrt{1-y^2}+z \sqrt{1-z^2}=2 x y z$.





Question 18

फलन को निम्नतम रूप में लिखिए :
$\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right) ;-\frac{\pi}{4}<x<\frac{3 \pi}{4}$






Question 19

यदि $\sin (\pi \cos \theta)=\cos (\pi \sin \theta)$, तो सिद्ध कीजिए :

$\theta=\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)$







Question 20

सिद्ध कीजिए कि :
$\tan ^{-1}\left(\frac{a-b}{1+a b}\right)+\tan ^{-1}\left(\frac{b-c}{1+b c}\right)+\tan ^{-1} c=\tan ^{-1} a$





Question 21

यदि $\cos ^{-1} \frac{x}{2}+\cos ^{-1} \frac{y}{3}=\theta$ तो सिद्ध कीजिए कि
$9 x^2-12 x y \cos \theta+4 y^2=36 \sin ^2 \theta$.
Sol :




Question 22

22. यदि $\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$, तो सिद्ध कीजिए कि :
$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2 x y}{a b} \cos \alpha=\sin ^2 \alpha$
Sol :





Question 23

निम्न समीकरणों को हल कीजिए :
23. $\sin ^{-1} x+\sin ^{-1}(1-x) \Rightarrow \cos ^{-1} x$.




Question 24

24. $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$.





Question 25

25. $\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x, x>0$.








Question 26

26. $\tan ^{-1} \frac{2 x}{x^2-1}+\cot ^{-1} \frac{x^2-1}{2 x}+\frac{4 \pi}{3}=0$.









Question 27

27. $\tan ^{-1}(x+1)-\tan ^{-1}(x-1)=\cot ^{-1} 2$.






Question 28

28. $\tan \left(\cos ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{1}{2}\right)$.






Question 29

29. $\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}$.








Question 30

$\tan ^{-1} \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\alpha$









Question 31

31. यदि $\sin ^{-1} x+\tan ^{-1} x=\frac{\pi}{2}$ तो सिद्ध कीजिए $2 x^2+1=\sqrt{5}$







Question 32

$\begin{aligned} & \sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3} \\ & \cos ^{-1} x-\cos ^{-1} y=\frac{\pi}{3}\end{aligned}$




Question 33

यदि $\frac{m \tan (\alpha-\theta)}{\cos ^2 \theta}=\frac{n \tan \theta}{\cos ^2(\alpha-\theta)}$ तो सिद्ध कीजिए कि
$2 \theta=\alpha-\left[\tan ^{-1}\left(\frac{n-m}{n+m}\right) \tan \alpha\right]$





Question 34

34. $\left(\sec ^{-1} \frac{x}{a}-\sec ^{-1} \frac{x}{b}\right)=\left(\sec ^{-1} b-\sec ^{-1} a\right), x$ का मान निकालिए।






Question 35

35. $\tan ^{-1}(x-1)+\tan ^{-1} x+\tan ^{-1}(x+1)=\tan ^{-1} 3 x$ को हल कीजिए।




















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