Question
What is the value of : $$\cos \left[ {{{\tan }^{ - 1}}\left\{ {\tan \left( {\frac{{15\pi }}{4}} \right)} \right\}} \right]\,?$$
A.
$$ - \frac{1}{{\sqrt 2 }}$$
B.
$$0$$
C.
$$ \frac{1}{{\sqrt 2 }}$$
D.
$$ \frac{1}{{2 \sqrt 2 }}$$
Answer :
$$ \frac{1}{{\sqrt 2 }}$$
Solution :
The given trigonometric expression is :
$$\eqalign{
& \cos \left[ {{{\tan }^{ - 1}}\left\{ {\tan \left( {\frac{{15\pi }}{4}} \right)} \right\}} \right] \cr
& = \cos \left[ {{{\tan }^{ - 1}}\left\{ {\tan \left( {4\pi - \frac{\pi }{4}} \right)} \right\}} \right] \cr
& = \cos \left[ {{{\tan }^{ - 1}}\left\{ { - \tan \frac{\pi }{4}} \right\}} \right] = \cos \left[ {{{\tan }^{ - 1}}\tan \left( {\frac{{ - \pi }}{4}} \right)} \right] \cr} $$
Since, $${\tan ^{ - 1}}\theta $$ is defined for $$\frac{{ - \pi }}{2} < \theta < \frac{{ - \pi }}{2}$$
$$\eqalign{
& = \cos \left( {\frac{{ - \pi }}{4}} \right) \cr
& \cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}\,\,\left[ {{\text{since, }}\cos \left( { - \theta } \right) = \cos \theta } \right] \cr} $$