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Question: How do you verify \( {\tan ^5}x = {\tan ^3}x{\sec ^2}x - {\tan ^3}x \) ?...

How do you verify tan⁡5x=tan⁡3xsec⁡2x−tan⁡3x{\tan ^5}x = {\tan ^3}x{\sec ^2}x - {\tan ^3}x ?

Explanation

Solution

Hint : Here, you are given a combination of trigonometric ratios: tangent and secant and are asked to prove the identity. What you need to do here is use the trigonometric identities such as the relation between secant and tangent of an angle, that is tan⁡2x=sec⁡2x−1{\tan ^2}x = {\sec ^2}x - 1 . Also, in order to proceed, you need to rearrange and write the left-hand side in a unique way which will help you to prove this identity, only if you are looking to bring LHS equal to RHS. If you are thinking of making RHS equal to LHS, then also you need to do some manipulation and write it in a unique way, so that you can apply the trigonometric properties known to you.

Complete step by step solution:
Let us first consider the left-hand side of the equation, that is tan⁡5x{\tan ^5}x . The quantity is tan⁡x\tan x raised to the power 55 , what you need to do here is split the power 55 into 22 and 33 , rewrite it as tan⁡5x=(tan⁡3x)(tan⁡2x){\tan ^5}x = \left( {{{\tan }^3}x} \right)\left( {{{\tan }^2}x} \right) . Now, tan⁡2x{\tan ^2}x can be expressed in terms of sec⁡x\sec x according to this property, tan⁡2x=sec⁡2x−1{\tan ^2}x = {\sec ^2}x - 1 . We will substitute this value of tan⁡2x{\tan ^2}x in the above modified equation. We get,
tan⁡5x=(tan⁡3x)(tan⁡2x) tan⁡5x=(tan⁡3x)(sec⁡2x−1) tan⁡5x=tan⁡3xsec⁡2x−tan⁡3x   {\tan ^5}x = \left( {{{\tan }^3}x} \right)\left( {{{\tan }^2}x} \right) \\\ {\tan ^5}x = \left( {{{\tan }^3}x} \right)\left( {{{\sec }^2}x - 1} \right) \\\ {\tan ^5}x = {\tan ^3}x{\sec ^2}x - {\tan ^3}x \;
Hence proved. You could have gone in the reverse way too by first considering the right-hand side to be manipulated. Let us go through that way also.
tan⁡3xsec⁡2x−tan⁡3x =tan⁡3x(sec⁡2x−1) =tan⁡3x(tan⁡2x) =tan⁡5x   {\tan ^3}x{\sec ^2}x - {\tan ^3}x \\\ = {\tan ^3}x\left( {{{\sec }^2}x - 1} \right) \\\ = {\tan ^3}x\left( {{{\tan }^2}x} \right) \\\ = {\tan ^5}x \;
From both sides, we have achieved the same result.
Hence proved tan⁡5x=tan⁡3xsec⁡2x−tan⁡3x{\tan ^5}x = {\tan ^3}x{\sec ^2}x - {\tan ^3}x , identity verified.

Note : The main idea is to arrange and rewrite the given equation so that you can apply the trigonometric identities in order to prove the identity which is asked to you. You need to memorize all the basic and fundamental properties of trigonometry such as the one which is used here, tan⁡2x=sec⁡2x−1{\tan ^2}x = {\sec ^2}x - 1 . These identities help you to simplify the given problem.