Question

Prove that $\dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = 2\cos ecA$

Answer 1

Hint:
Here, we will prove the relation between the given trigonometric Identity. We will first simplify the terms on the left-hand side of the given trigonometric relation by multiplying with their conjugate. Then by subtracting the simplified expression and by using trigonometric ratio, we will prove the given trigonometric relation.

Formula Used:
We will use the following formula:
1) Trigonometric Identity: $1 - {\sec ^2}A = - {\tan ^2}A$
2) Trigonometric Ratio: $\sec A = \dfrac{1}{{\cos A}}$
3) Trigonometric Ratio: $\tan A = \dfrac{{\sin A}}{{\cos A}}$

Complete step by step solution:
We are given to prove that $\dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = 2\cos ecA$
Now, considering $\dfrac{{\tan A}}{{1 + \sec A}}$ and multiplying with its conjugate, we get
$\dfrac{{\tan A}}{{1 + \sec A}} = \dfrac{{\tan A}}{{1 + \sec A}} \times \dfrac{{1 - \sec A}}{{1 - \sec A}}$
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} = \dfrac{{\tan A\left( {1 - \sec A} \right)}}{{\left( {1 + \sec A} \right)\left( {1 - \sec A} \right)}}$
By multiplying the terms in the parenthesis, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} = \dfrac{{\tan A - \tan A\sec A}}{{1 - {{\sec }^2}A + \sec A - \sec A}}$
By cancelling the same terms in the denominator, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} = \dfrac{{\tan A\left( {1 - \sec A} \right)}}{{1 - {{\sec }^2}A}}$………………………………………………………………………………….$\left( 1 \right)$
Now, considering $\dfrac{{\tan A}}{{1 - \sec A}}$ and multiplying with its conjugate, we get
$\dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{\tan A}}{{1 - \sec A}} \times \dfrac{{1 + \sec A}}{{1 + \sec A}}$
$ \Rightarrow \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{\tan A\left( {1 + \sec A} \right)}}{{\left( {1 - \sec A} \right)\left( {1 + \sec A} \right)}}$
By multiplying the terms in the parenthesis, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{\tan A\left( {1 + \sec A} \right)}}{{1 - {{\sec }^2}A + \sec A - \sec A}}$
By cancelling the same terms in the denominator, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{\tan A\left( {1 + \sec A} \right)}}{{1 - {{\sec }^2}A}}$……………………………………………………………………………$\left( 2 \right)$
By subtracting equation $\left( 2 \right)$ from equation $\left( 1 \right)$ , we get
$\dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{\tan A\left( {1 - \sec A} \right)}}{{1 - {{\sec }^2}A}} - \dfrac{{\tan A\left( {1 + \sec A} \right)}}{{1 - {{\sec }^2}A}}$
Factoring out common term, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{\tan A\left[ {\left( {1 - \sec A} \right) - \left( {1 + \sec A} \right)} \right]}}{{1 - {{\sec }^2}A}}$
By simplifying the equation, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{\tan A\left[ {1 - \sec A - 1 - \sec A} \right]}}{{1 - {{\sec }^2}A}}$
By adding and subtracting like terms, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{\tan A\left[ { - 2\sec A} \right]}}{{ - {{\tan }^2}A}}$
By cancelling the like terms, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{ - 2\sec A}}{{ - \tan A}}$
By rewriting in terms of sine and cosine ratios, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{{2\dfrac{1}{{\cos A}}}}{{\dfrac{{\sin A}}{{\cos A}}}}$
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{2}{{\cos A}} \times \dfrac{{\cos A}}{{\sin A}}$
By cancelling like terms, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = \dfrac{2}{{\sin A}}$
By using reciprocal trigonometric ratio, we get
$ \Rightarrow \dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = 2\cos ecA$

Therefore, $\dfrac{{\tan A}}{{1 + \sec A}} - \dfrac{{\tan A}}{{1 - \sec A}} = 2\cos ecA$ is proved.

Note:
Trigonometric equation is defined as an equation involving the trigonometric ratios. Trigonometric identity is an equation which is always true for all the variables. We need to keep in mind that the trigonometric ratio and the co-trigonometric ratio is always reciprocal to each other. Trigonometric ratios are used to find the relationships between the sides of a right angle triangle. Conjugate is a term where the sign is changed between two terms.