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Question: Prove the identity\(\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = 1 - 2\sec \...

Prove the identitysec⁡θ−tan⁡θsec⁡θ+tan⁡θ=1−2sec⁡θtan⁡θ+2tan⁡2θ\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = 1 - 2\sec \theta \tan \theta + 2{\tan ^2}\theta .

Explanation

Solution

Hint: - Here we go through by applying the properties of rationalization, first apply the rationalization in the left hand side and then apply the trigonometric identities to prove the result given in the right hand side.

“Complete step-by-step answer:”
Given sec⁡θ−tan⁡θsec⁡θ+tan⁡θ=1−2sec⁡θtan⁡θ+2tan⁡2θ\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = 1 - 2\sec \theta \tan \theta + 2{\tan ^2}\theta
Let us assume the function on the left hand side L.H.S. i.e. sec⁡θ−tan⁡θsec⁡θ+tan⁡θ\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} and the function that is on the right hand side R.H.S. i.e. 1−2sec⁡θtan⁡θ+2tan⁡2θ1 - 2\sec \theta \tan \theta + 2{\tan ^2}\theta .

Let us consider the L.H.S.
⇒sec⁡θ−tan⁡θsec⁡θ+tan⁡θ\Rightarrow \dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} Here we apply the rationalization rule to make in the form of trigonometric identities.
I.e. (sec⁡θ−tan⁡θsec⁡θ+tan⁡θ)×(sec⁡θ−tan⁡θsec⁡θ−tan⁡θ)\left( {\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }}} \right) \times \left( {\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta - \tan \theta }}} \right) as we know in the rationalization we multiply both top and bottom by the conjugate of the denominator.
⇒(sec⁡θ−tan⁡θ)2sec⁡2θ−tan⁡2θ\Rightarrow \dfrac{{{{\left( {\sec \theta - \tan \theta } \right)}^2}}}{{{{\sec }^2}\theta - {{\tan }^2}\theta }} As we know sec⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1 so we can write it as,
⇒(sec⁡θ−tan⁡θ)21\Rightarrow \dfrac{{{{\left( {\sec \theta - \tan \theta } \right)}^2}}}{1}
⇒(sec⁡θ−tan⁡θ)2=sec⁡2θ+tan⁡2θ−2sec⁡θtan⁡θ\Rightarrow {\left( {\sec \theta - \tan \theta } \right)^2} = {\sec ^2}\theta + {\tan ^2}\theta - 2\sec \theta \tan \theta As we know by algebraic formula (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab
⇒(1+tan⁡2θ)+tan⁡2θ−2sec⁡θtan⁡θ\Rightarrow \left( {1 + {{\tan }^2}\theta } \right) + {\tan ^2}\theta - 2\sec \theta \tan \theta ∵\because (sec⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1)
⇒1−2sec⁡θtan⁡θ+2tan⁡2θ\Rightarrow 1 - 2\sec \theta \tan \theta + 2{\tan ^2}\theta
Here we can see that the L.H.S is equal to the R.H.S.
Hence, proved.

Note:- Whenever we face such a type of question in which the conjugate of numerator is given in denominator then the key concept for solving the question is always try to start with applying the rationalization rule and then for proving this question we apply the trigonometry identity .