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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 $\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. $\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }}$ and the function that is on the right hand side R.H.S. i.e. $1 - 2\sec \theta \tan \theta + 2{\tan ^2}\theta $.

Let us consider the L.H.S.

$ \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. $\left( {\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }}} \right) \times \left( {\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta - \tan \theta }}} \right)$ as we know in the rationalization we multiply both top and bottom by the conjugate of the denominator.

$ \Rightarrow \dfrac{{{{\left( {\sec \theta - \tan \theta } \right)}^2}}}{{{{\sec }^2}\theta - {{\tan }^2}\theta }}$ As we know ${\sec ^2}\theta - {\tan ^2}\theta = 1$ so we can write it as,

$ \Rightarrow \dfrac{{{{\left( {\sec \theta - \tan \theta } \right)}^2}}}{1}$

$ \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 ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$

$ \Rightarrow \left( {1 + {{\tan }^2}\theta } \right) + {\tan ^2}\theta - 2\sec \theta \tan \theta $ $\because $(${\sec ^2}\theta - {\tan ^2}\theta = 1$)

$ \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 .

“Complete step-by-step answer:”

Given $\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. $\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }}$ and the function that is on the right hand side R.H.S. i.e. $1 - 2\sec \theta \tan \theta + 2{\tan ^2}\theta $.

Let us consider the L.H.S.

$ \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. $\left( {\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }}} \right) \times \left( {\dfrac{{\sec \theta - \tan \theta }}{{\sec \theta - \tan \theta }}} \right)$ as we know in the rationalization we multiply both top and bottom by the conjugate of the denominator.

$ \Rightarrow \dfrac{{{{\left( {\sec \theta - \tan \theta } \right)}^2}}}{{{{\sec }^2}\theta - {{\tan }^2}\theta }}$ As we know ${\sec ^2}\theta - {\tan ^2}\theta = 1$ so we can write it as,

$ \Rightarrow \dfrac{{{{\left( {\sec \theta - \tan \theta } \right)}^2}}}{1}$

$ \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 ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$

$ \Rightarrow \left( {1 + {{\tan }^2}\theta } \right) + {\tan ^2}\theta - 2\sec \theta \tan \theta $ $\because $(${\sec ^2}\theta - {\tan ^2}\theta = 1$)

$ \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 .

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