Question

The value of $\cos \left[{\tan }^{-1}\left(tan\frac{15\pi }{4}\right)\right]$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $-\frac{1}{\sqrt{2}}$
• C. 1
• D. none of these

Answer 1

Answer: (A)

$ta{n}^{-1}\left(tanx\right)=x$ if $-\frac{\pi }{2}<x<\frac{\pi }{2}$
$tan\frac{15\pi }{4}=tan\left(4\pi -\frac{\pi }{4}\right)=tan\left(-\frac{\pi }{4}\right)$
$\therefore ta{n}^{-1}\left(tan\frac{15\pi }{4}\right)ta{n}^{-1}\left(tan\left(-\frac{\pi }{4}\right)\right)=-\frac{\pi }{4}$
$\therefore cos\left[ta{n}^{-1}\left(tan\frac{15\pi }{4}\right)\right]=cos\left(-\frac{\pi }{4}\right)$
$=cos\frac{\pi }{4}$
$=\frac{1}{\sqrt{2}}$