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Question: Prove: \[\cos e{c^2}\theta + {\sec ^2}\theta = {\sec ^2}\theta .\cos e{c^2}\theta \]?...

Prove: cos⁡ec2θ+sec⁡2θ=sec⁡2θ.cos⁡ec2θ\cos e{c^2}\theta + {\sec ^2}\theta = {\sec ^2}\theta .\cos e{c^2}\theta ?

Explanation

Solution

Hint : To prove this type of question, we need to use some of the basic trigonometric identities and formulae. To prove the given result, we have to make right side of the equation equals to left side of the equation using trigonometric identities such as sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 , sec⁡x=1cos⁡x\sec x = \dfrac{1}{{\cos x}} and cos⁡ec(x)=1sin⁡(x)\cos ec\left( x \right) = \dfrac{1}{{\sin \left( x \right)}} . To prove the desired result, we have to proceed with the left hand side of the equation and simplify and manipulate it to get the right side of the equation.

Complete step-by-step answer :
In the given question, we are required to prove the given trigonometric equation with the help of trigonometric formulae and identities.
So, we proceed with the left side of the equation,
cos⁡ec2θ+sec⁡2θ\cos e{c^2}\theta + {\sec ^2}\theta
Using the basic trigonometric formulae sec⁡x=1cos⁡x\sec x = \dfrac{1}{{\cos x}} and cos⁡ec(x)=1sin⁡(x)\cos ec\left( x \right) = \dfrac{1}{{\sin \left( x \right)}} , we get,
=1sin⁡2θ+1cos⁡2θ\dfrac{1}{{{{\sin }^2}\theta }} + \dfrac{1}{{{{\cos }^2}\theta }}
Taking LCM of the denominators, we get,
=cos⁡2θ+sin⁡2θsin⁡2θ.cos⁡2θ= \dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\sin }^2}\theta .{{\cos }^2}\theta }}
Applying the basic trigonometric identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 , we get,
=1sin⁡2θ.cos⁡2θ= \dfrac{1}{{{{\sin }^2}\theta .{{\cos }^2}\theta }}
Separating the denominators, we get,
=1sin⁡2θ.1cos⁡2θ= \dfrac{1}{{{{\sin }^2}\theta }}.\dfrac{1}{{{{\cos }^2}\theta }}
Using the same trigonometric formulae sec⁡x=1cos⁡x\sec x = \dfrac{1}{{\cos x}} and cos⁡ec(x)=1sin⁡(x)\cos ec\left( x \right) = \dfrac{1}{{\sin \left( x \right)}} again, we get,
=cos⁡ec2θ.sec⁡2θ= \cos e{c^2}\theta .{\sec ^2}\theta
Hence, the given trigonometric equation is proved.
So, the correct answer is “ cos⁡ec2θ.sec⁡2θ\cos e{c^2}\theta .{\sec ^2}\theta ”.

Note : Such questions that require us to prove a trigonometric equation require thorough knowledge of trigonometric functions, formulae and trigonometric identities. Algebraic operations and rules such as taking LCM of denominators and transposition of terms to isolate a specific term, play a significant role in the solution of such problems.