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Question: If \(\tan \left( {\pi \cos \theta } \right) = \cot \left( {\pi \sin \theta } \right)\) then prove th...

If tan⁡(πcos⁡θ)=cot⁡(πsin⁡θ)\tan \left( {\pi \cos \theta } \right) = \cot \left( {\pi \sin \theta } \right) then prove that cos⁡(θ−π4)=±122\cos \left( {\theta - \dfrac{\pi }{4}} \right) = \pm \dfrac{1}{{2\sqrt 2 }}.

Explanation

Solution

1. Change the R.H.S trigonometric ratio in its complementary ratio, by using identity:
cot⁡θ=tan⁡(±π2−θ)\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⁡θ=tan⁡α\tan \theta = \tan \alpha
θ=nπ+α\theta = n\pi + \alpha
Where, α∈[−π2,π2]\alpha \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] and n∈Zn \in Z.
3. Then try to make cos⁡(θ−π4)\cos \left( {\theta - \dfrac{\pi }{4}} \right) on the L.H.S.

Complete step-by-step answer:
It is given that,
tan⁡(πcos⁡θ)=cot⁡(πsin⁡θ)\tan \left( {\pi \cos \theta } \right) = \cot \left( {\pi \sin \theta } \right)
⇒tan⁡(πcos⁡θ)=tan⁡(±π2−πsin⁡θ) (∵tan⁡(±π2−θ)=cot⁡θ)\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⁡θ=tan⁡α\tan \theta = \tan \alpha is given by:
θ=nπ+α\theta = n\pi + \alpha
Where, α∈[−π2,π2]\alpha \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] and n∈Zn \in Z.
So, here in the above equation we get:
⇒πcos⁡θ=nπ+(±π2−πsin⁡θ).................................[1]\Rightarrow \pi \cos \theta = n\pi + \left( { \pm \dfrac{\pi }{2} - \pi \sin \theta } \right).................................[1]
Where,
−π2⩽π2−sin⁡θ⩽π2 or −π2⩽−π2−sin⁡θ⩽π2\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}
⇒−π⩽−πsin⁡θ⩽0 or 0⩽ - πsinθ⩽π ⇒0⩽sin⁡θ⩽1 or - 1⩽sinθ⩽0 ⇒ - 1⩽sinθ⩽1 ⇒θ∈[−π2,π2]  \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 π\pi , we get:
⇒cos⁡θ=n±12−sin⁡θ ⇒cos⁡θ+sin⁡θ=n±12  \Rightarrow \cos \theta = n \pm \dfrac{1}{2} - \sin \theta \\\ \Rightarrow \cos \theta + \sin \theta = n \pm \dfrac{1}{2} \\\
For θ∈[−π2,π2]\theta \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]the value of cos⁡θ+sin⁡θ\cos \theta + \sin \theta lies between −2- \sqrt 2 and 2\sqrt 2 .
Therefore, n can only be 0.
⇒cos⁡θ+sin⁡θ=±12\Rightarrow \cos \theta + \sin \theta = \pm \dfrac{1}{2}
Dividing both sides by 2\sqrt 2 , we get:
⇒12cos⁡θ+12sin⁡θ=±122 ⇒cos⁡π4cos⁡θ+sin⁡π4sin⁡θ=±122 [cosπ4=sin⁡π4=12]  \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⁡Acos⁡B+sin⁡Asin⁡B=cos⁡(A−B)\cos A\cos B + \sin A\sin B = \cos (A - B)we get,
⇒cos⁡(π4−θ)=±122\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 θ\theta was not given that’s why when we calculated we got θ∈[−π2,π2]\theta \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right].
Since, the value of tan⁡θ\tan \theta is positive in both I quadrant and II quadrant, therefore it is important to take ±\pm sign in the identity:
cot⁡θ=tan⁡(±π2−θ)\cot \theta = \tan \left( { \pm \dfrac{\pi }{2} - \theta } \right).