inverse trigonometric functions

 $If\ \sin\left(\cot^{-1}\left(x+1\right)\right)=\cos\left(\tan^{-1}x\right),\ then\ find\ the\ value\ of\ x$  

•  $-\frac{1}{2}$  

•  $\frac{1}{2}$  

•  $0$  

•  $\frac{9}{4}$  

 $\tan^{-1}\ A\ =\sin^{-1}\left(\frac{5}{13}\right)+\cos^{-1}\left(\frac{3}{5}\right).\ Here\ A\ refers\ to$  

• 63/16

• 63/65

• 12/13

• 25/63

 $\ \ Reason\left(R\right):\tan^{-1}\left(\frac{\left(\sqrt{1+x}-\sqrt{1-x}\right)}{\left(\sqrt{1+x}+\sqrt{1-x}\right)}\right)\$   $-\frac{1}{\sqrt{2}}\le x\le1$ 

Assertion(A)   
 $\ \ \ \ \ \ \ \cot^{-1}\left(\frac{\sqrt{\left(1+\sin x\right)}+\sqrt{\left(1-\sin x\right)}}{\sqrt{\left(1+\sin x\right)}-\sqrt{\left(1-\sin x\right)}}\right)$ 

 $x\in\left(\ 0,\ \frac{\pi}{4}\right)$  

• Both A & R are correct ; R is the correct explanation of A

• Both A & R are correct ; R is the not correct explanation of A

• A is correct ; R is incorrect

• R is correct ; A is incorrect

 $The\ domain\ ofthe\ function\ \ \ y=\sin^{-1}\left(-x^2\right)$  

• [ 0 , 1]

• (0 , 1]

• [ -1 , 1]

•  $\phi$  

 $\cos^{-1\ }x>\ \sin^{-1\ }x\ ,\ then$  

•  $\frac{1}{\sqrt{2}}<x\ <1$  

•  $0\le\ x\ <\frac{1}{\sqrt{2}}$  

•  $-1\le x<\frac{1}{\sqrt{2}}$  

•  $x>0$  

 $The\ value\ of\ \ \sin\left[2\ \sin^{-1}\left(0.6\right)\right]$  

• 0.48

• 0.96

• 1.2

• sin 1.2

• $A$

• $B$

• $C$

• $D$

• A

• B

• C

• D

 $2\ \tan^{-1}\ \frac{1}{5}\ +\ \sec^{-1}\ \frac{5\sqrt{2}}{7}+2\ \tan^{-1}\ \frac{1}{8}=$  

(a)  

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

•  $\frac{\left(4+\sqrt{7}\right)}{3}$  

•  $\frac{\left(4-\sqrt{7}\right)}{3}$