Question

The value of $\underset{x\to \frac{1}{\sqrt{2}}}{lim}\frac{x-\cos \left({\sin }^{-1}x\right)}{1-\tan \left({\sin }^{-1}x\right)}$ is

• A. $-\frac{1}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\sqrt{2}$
• D. $-\sqrt{2}$

Answer 1

Answer: (A)

$-\frac{1}{\sqrt{2}}$

$\underset{x\to \frac{1}{\sqrt{2}}}{lim}\frac{x-\cos \left(si{n}^{-1}x\right)}{1-\tan \left(si{n}^{-1}x\right)}$
Let ${\sin }^{-1}x=\theta$
Then,$x=\sin \theta$
Also, $x\to \frac{1}{\sqrt{2}}\Rightarrow \theta \to \frac{\pi }{4}$
So, $\underset{x\to \frac{1}{\sqrt{2}}}{lim}\frac{x-\cos \left({\sin }^{-1}x\right)}{1-\tan \left(si{n}^{-1}x\right)}=\underset{\theta \to \frac{\pi }{4}}{lim}\frac{\sin \theta -\cos \theta }{1-\tan \theta }$
$=\underset{\theta \to \frac{\pi }{4}}{lim}\frac{(\sin \theta -\cos \theta )}{(\cos \theta -\sin \theta )}\cos \theta$
$=\underset{\theta \to \frac{\pi }{4}}{lim}-\cos \theta =-\frac{1}{\sqrt{2}}$