# If trigonometric equation is given as $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$ then $\cos (\theta - \dfrac{\pi }{4})$ is equal to

\[

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 \\

\]

Answer

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

Complete step-by-step solution -

Given that $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$

Now, proceeding further with the given equation

$ \Rightarrow \tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$

As we know that $\left[ {\cot A = \tan (\dfrac{\pi }{2} - A)} \right]$

Using the above formula in the given equation, we get

$

\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 \[\dfrac{{\sqrt 2 }}{{\sqrt 2 }}\] to from a cosine formula

$

\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 that$\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4}$ , substituting this in the above formula

$ \Rightarrow \sqrt 2 (\cos \theta \cos \dfrac{\pi }{4} + \sin \theta \sin \dfrac{\pi }{4}) = \pm \dfrac{1}{2}$

As we know that$\left[ {\cos (A - B) = \cos A\cos B + \sin A\sin B} \right]$

$ \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.

Complete step-by-step solution -

Given that $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$

Now, proceeding further with the given equation

$ \Rightarrow \tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$

As we know that $\left[ {\cot A = \tan (\dfrac{\pi }{2} - A)} \right]$

Using the above formula in the given equation, we get

$

\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 \[\dfrac{{\sqrt 2 }}{{\sqrt 2 }}\] to from a cosine formula

$

\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 that$\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4}$ , substituting this in the above formula

$ \Rightarrow \sqrt 2 (\cos \theta \cos \dfrac{\pi }{4} + \sin \theta \sin \dfrac{\pi }{4}) = \pm \dfrac{1}{2}$

As we know that$\left[ {\cos (A - B) = \cos A\cos B + \sin A\sin B} \right]$

$ \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.

Last updated date: 18th Sep 2023

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