Question

Solve $\dfrac{{\cot A - 1}}{{2 - {{\sec }^2}A}} = \dfrac{{\cot A}}{{1 + \tan A}}$ .

Answer 1

Hint: In this question, we are given two terms one on the left- hand side and the other on the right- hand side and we have to prove the left-hand side is equal to the right-hand side.
Convert $\cot A$ and $\tan A$ into $\sin A$ and $\cos A$ forms on both sides, and then solve both sides separately until they are equal to each other.

 Formula to be used:
$\cot A = \dfrac{{\cos A}}{{\sin A}}$
$\sec A = \dfrac{1}{{\cos A}}$
$\cos 2A = \left( {2{{\cos }^2}A - 1} \right)$
$\cos 2A = {\cos ^2}A - {\sin ^2}A$
${a^2} - {b^2} = \left( {a + b} \right)(a - b)$

Complete step-by-step answer:
Given equation $\dfrac{{\cot A - 1}}{{2 - {{\sec }^2}A}} = \dfrac{{\cot A}}{{1 + \tan A}}$ .
To prove the left-hand side is equal to the right-hand side.
First, consider left-hand side, $\dfrac{{\cot A - 1}}{{2 - {{\sec }^2}A}}$ , and convert it into $\sin A$ and $\cos A$ form, we get, $\dfrac{{\dfrac{{\cos A}}{{\sin A}} - 1}}{{2 - \dfrac{1}{{{{\cos }^2}A}}}}$ .
Taking least common multiple, and solving, $\dfrac{{\left( {\cos A - \sin A} \right){{\cos }^2}A}}{{\left( {2{{\cos }^2}A - 1} \right)\sin A}}$ .
Now, we can write $\cos 2A = \left( {2{{\cos }^2}A - 1} \right)$ in the denominator, we get, $\dfrac{{\left( {\cos A - \sin A} \right){{\cos }^2}A}}{{\cos 2A\sin A}}$ .
Now, write $\cos 2A = {\cos ^2}A - {\sin ^2}A$ , we get, $\dfrac{{\left( {\cos A - \sin A} \right){{\cos }^2}A}}{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\sin A}}$ .
Applying the identity ${a^2} - {b^2} = \left( {a + b} \right)(a - b)$ , we get, $\dfrac{{\left( {\cos A - \sin A} \right){{\cos }^2}A}}{{\left( {\cos A - \sin A} \right)\left( {\cos A + \sin A} \right)\sin A}}$ .
On solving, we get, $\dfrac{{{{\cos }^2}A}}{{\left( {\cos A + \sin A} \right)\sin A}}$ .
Now, consider right-hand side, $\dfrac{{\cot A}}{{1 + \tan A}}$ , convert it into $\sin A$ and $\cos A$ form, we get, $\dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{1 + \dfrac{{\sin A}}{{\cos A}}}}$ .
Taking least common multiple, and solving, $\dfrac{{{{\cos }^2}A}}{{\sin A(\cos A + \sin A)}}$ , which is equal to the left-hand side.
Thus, $\dfrac{{{{\cos }^2}A}}{{\sin A(\cos A + \sin A)}} = \dfrac{{{{\cos }^2}A}}{{\sin A(\cos A + \sin A)}}$ i.e., left-hand side is equal to right-hand side.
Hence, proved.

Note: Alternate way to solve the question is, consider left-hand side, $\dfrac{{\cot A - 1}}{{2 - {{\sec }^2}A}}$ and convert each term into $\tan A$ form, we get, $\dfrac{{\dfrac{1}{{\tan A}} - 1}}{{2 - (1 + {{\tan }^2}A)}}$ , taking least common multiple and solving, we get, $\dfrac{{\dfrac{{1 - \tan A}}{{\tan A}}}}{{1 - {{\tan }^2}A}}$ , which gives, $\dfrac{{1 - \tan A}}{{\tan A(1 - {{\tan }^2}A)}}$ . Now, using the identity, ${a^2} - {b^2} = \left( {a + b} \right)(a - b)$ , $\dfrac{{1 - \tan A}}{{\tan A(1 - \tan A)(1 + \tan A)}}$ which gives $\dfrac{1}{{\tan A\left( {1 + \tan A} \right)}}$ which can also be written as $\dfrac{{\cot A}}{{\left( {1 + \tan A} \right)}}$ , hence, proved.
One must remember the trigonometric identities, to solve such types of questions.
In questions with trigonometric functions, if you don’t find a way of solving the question, then convert both sides into $\cos A$ and $\sin A$ form and simplify.