If trigonometric equation is given as $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$ then $\cos (\theta - \dfrac{\pi }{4})$ is equal to
\[
A.{\text{ }} \pm \dfrac{1}{{2\sqrt 2 }} \\
B.{\text{ }} \pm \dfrac{1}{{\sqrt 2 }} \\
C.{\text{ }} \pm \sqrt 2 \\
D.{\text{ }} \pm 2\sqrt 2 \\
\]
Last updated date: 24th Mar 2023
•
Total views: 305.1k
•
Views today: 2.83k
Answer
305.1k+ views
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
