Question

If $\tan \left( {\pi \cos \theta } \right) = \cot \left( {\pi \sin \theta } \right)$ then prove that $\cos \left( {\theta - \dfrac{\pi }{4}} \right) = \pm \dfrac{1}{{2\sqrt 2 }}$.

Answer 1

Hint: 1. Change the R.H.S trigonometric ratio in its complementary ratio, by using identity:
$\cot \theta = \tan \left( { \pm \dfrac{\pi }{2} - \theta } \right)$.
2. Then compare the phases of both L.H.S and R.H.S, by using identity:
When, $\tan \theta = \tan \alpha $
$\theta = n\pi + \alpha $
Where, $\alpha \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ and $n \in Z$.
3. Then try to make $\cos \left( {\theta - \dfrac{\pi }{4}} \right)$ on the L.H.S.

Complete step-by-step answer:
It is given that,
$\tan \left( {\pi \cos \theta } \right) = \cot \left( {\pi \sin \theta } \right)$
$ \Rightarrow \tan (\pi \cos \theta ) = \tan \left( { \pm \dfrac{\pi }{2} - \pi \sin \theta } \right){\text{ }}\left( {\because \tan \left( { \pm \dfrac{\pi }{2} - \theta } \right) = \cot \theta } \right)$
Now. As we know the general solution of the equation:
$\tan \theta = \tan \alpha $ is given by:
$\theta = n\pi + \alpha $
Where, $\alpha \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ and $n \in Z$.
So, here in the above equation we get:
$ \Rightarrow \pi \cos \theta = n\pi + \left( { \pm \dfrac{\pi }{2} - \pi \sin \theta } \right).................................[1]$
Where,
$\dfrac{{ - \pi }}{2} \leqslant \dfrac{\pi }{2} - \sin \theta \leqslant \dfrac{\pi }{2}{\text{ or }}\dfrac{{ - \pi }}{2} \leqslant - \dfrac{\pi }{2} - \sin \theta \leqslant \dfrac{\pi }{2}$
$
   \Rightarrow - \pi \leqslant - \pi \sin \theta \leqslant 0{\text{ or 0}} \leqslant {\text{ - }}\pi {\text{sin}}\theta \leqslant \pi \\
   \Rightarrow 0 \leqslant \sin \theta \leqslant 1{\text{ or - 1}} \leqslant {\text{sin}}\theta \leqslant {\text{0}} \\
   \Rightarrow {\text{ - 1}} \leqslant {\text{sin}}\theta \leqslant {\text{1}} \\
   \Rightarrow \theta \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] \\
 $
Now, dividing equation 1 by $\pi $, we get:
$
   \Rightarrow \cos \theta = n \pm \dfrac{1}{2} - \sin \theta \\
   \Rightarrow \cos \theta + \sin \theta = n \pm \dfrac{1}{2} \\
 $
For $\theta \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$the value of $\cos \theta + \sin \theta $lies between $ - \sqrt 2 $ and $\sqrt 2 $.
Therefore, n can only be 0.
$ \Rightarrow \cos \theta + \sin \theta = \pm \dfrac{1}{2}$
Dividing both sides by $\sqrt 2 $, we get:
$
   \Rightarrow \dfrac{1}{{\sqrt 2 }}\cos \theta + \dfrac{1}{{\sqrt 2 }}\sin \theta = \pm \dfrac{1}{{2\sqrt 2 }} \\
   \Rightarrow \cos \dfrac{\pi }{4}\cos \theta + \sin \dfrac{\pi }{4}\sin \theta = \pm \dfrac{1}{{2\sqrt 2 }}{\text{ [cos}}\dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}] \\
 $
Using the identity:
$\cos A\cos B + \sin A\sin B = \cos (A - B)$we get,
$ \Rightarrow \cos \left( {\dfrac{\pi }{4} - \theta } \right) = \pm \dfrac{1}{{2\sqrt 2 }}$
Hence, proved.

Note: Here, in the above question the range for $\theta $ was not given that’s why when we calculated we got $\theta \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$.
Since, the value of $\tan \theta $ is positive in both I quadrant and II quadrant, therefore it is important to take $ \pm $ sign in the identity:
$\cot \theta = \tan \left( { \pm \dfrac{\pi }{2} - \theta } \right)$.