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Question: If trigonometric equation is given as \(\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )\) then \(\...

If trigonometric equation is given as tan⁡(πcos⁡θ)=cot⁡(πsin⁡θ)\tan (\pi \cos \theta ) = \cot (\pi \sin \theta ) then cos⁡(θ−π4)\cos (\theta - \dfrac{\pi }{4}) is equal to

A. ±122 B. ±12 C. ±2 D. ±22  A.{\text{ }} \pm \dfrac{1}{{2\sqrt 2 }} \\\ B.{\text{ }} \pm \dfrac{1}{{\sqrt 2 }} \\\ C.{\text{ }} \pm \sqrt 2 \\\ D.{\text{ }} \pm 2\sqrt 2 \\\
Explanation

Solution

Hint- In order to solve this question we will use the simple trigonometric identities such as tan⁡(900−θ)=cot⁡θ\tan ({90^0} - \theta ) = \cot \theta and cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B.\cos (A - B) = \cos A\cos B + \sin A\sin B. So we will try to make the given term in this form to proceed further.

Complete step-by-step solution -
Given that tan⁡(πcos⁡θ)=cot⁡(πsin⁡θ)\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )
Now, proceeding further with the given equation
⇒tan⁡(πcos⁡θ)=cot⁡(πsin⁡θ)\Rightarrow \tan (\pi \cos \theta ) = \cot (\pi \sin \theta )
As we know that [cot⁡A=tan⁡(π2−A)]\left[ {\cot A = \tan (\dfrac{\pi }{2} - A)} \right]
Using the above formula in the given equation, we get
⇒tan⁡(πcos⁡θ)=tan⁡(±π2−πsin⁡θ) ⇒πcos⁡θ=±π2−πsin⁡θ ⇒π(cos⁡θ+sin⁡θ)=±π2 ⇒(cos⁡θ+sin⁡θ)=±12  \Rightarrow \tan (\pi \cos \theta ) = \tan ( \pm \dfrac{\pi }{2} - \pi \sin \theta ) \\\ \Rightarrow \pi \cos \theta = \pm \dfrac{\pi }{2} - \pi \sin \theta \\\ \Rightarrow \pi (\cos \theta + \sin \theta ) = \pm \dfrac{\pi }{2} \\\ \Rightarrow (\cos \theta + \sin \theta ) = \pm \dfrac{1}{2} \\\
Now, we will multiply LHS by 22\dfrac{{\sqrt 2 }}{{\sqrt 2 }} to from a cosine formula
⇒22(cos⁡θ+sin⁡θ)=±12 ⇒2(cos⁡θ2+sin⁡θ2)=±12  \Rightarrow \dfrac{{\sqrt 2 }}{{\sqrt 2 }}(\cos \theta + \sin \theta ) = \pm \dfrac{1}{2} \\\ \Rightarrow \sqrt 2 (\dfrac{{\cos \theta }}{{\sqrt 2 }} + \dfrac{{\sin \theta }}{{\sqrt 2 }}) = \pm \dfrac{1}{2} \\\
Since, we know thatcos⁡π4=12=sin⁡π4\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4} , substituting this in the above formula
⇒2(cos⁡θcos⁡π4+sin⁡θsin⁡π4)=±12\Rightarrow \sqrt 2 (\cos \theta \cos \dfrac{\pi }{4} + \sin \theta \sin \dfrac{\pi }{4}) = \pm \dfrac{1}{2}
As we know that[cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B]\left[ {\cos (A - B) = \cos A\cos B + \sin A\sin B} \right]
⇒cos⁡(θ−π4)=±122\Rightarrow \cos (\theta - \dfrac{\pi }{4}) = \pm \dfrac{1}{{2\sqrt 2 }}
Hence, the correct option is A

Note- In such type of questions starts solving from the complex side of the questions and tries to express every term in terms of sin and cosine or a variable which is easy to solve. To simplify these questions try to combine two terms to a single term using trigonometric formulas.