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Question: Solve \(\dfrac{{\cot A - 1}}{{2 - {{\sec }^2}A}} = \dfrac{{\cot A}}{{1 + \tan A}}\) ....

Solve cot⁡A−12−sec⁡2A=cot⁡A1+tan⁡A\dfrac{{\cot A - 1}}{{2 - {{\sec }^2}A}} = \dfrac{{\cot A}}{{1 + \tan A}} .

Explanation

Solution

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\cot A and tan⁡A\tan A into sin⁡A\sin A and cos⁡A\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=cos⁡Asin⁡A\cot A = \dfrac{{\cos A}}{{\sin A}}
sec⁡A=1cos⁡A\sec A = \dfrac{1}{{\cos A}}
cos⁡2A=(2cos⁡2A−1)\cos 2A = \left( {2{{\cos }^2}A - 1} \right)
cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = {\cos ^2}A - {\sin ^2}A
a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)(a - b)

Complete step-by-step answer:
Given equation cot⁡A−12−sec⁡2A=cot⁡A1+tan⁡A\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, cot⁡A−12−sec⁡2A\dfrac{{\cot A - 1}}{{2 - {{\sec }^2}A}} , and convert it into sin⁡A\sin A and cos⁡A\cos A form, we get, cos⁡Asin⁡A−12−1cos⁡2A\dfrac{{\dfrac{{\cos A}}{{\sin A}} - 1}}{{2 - \dfrac{1}{{{{\cos }^2}A}}}} .
Taking least common multiple, and solving, (cos⁡A−sin⁡A)cos⁡2A(2cos⁡2A−1)sin⁡A\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=(2cos⁡2A−1)\cos 2A = \left( {2{{\cos }^2}A - 1} \right) in the denominator, we get, (cos⁡A−sin⁡A)cos⁡2Acos⁡2Asin⁡A\dfrac{{\left( {\cos A - \sin A} \right){{\cos }^2}A}}{{\cos 2A\sin A}} .
Now, write cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = {\cos ^2}A - {\sin ^2}A , we get, (cos⁡A−sin⁡A)cos⁡2A(cos⁡2A−sin⁡2A)sin⁡A\dfrac{{\left( {\cos A - \sin A} \right){{\cos }^2}A}}{{\left( {{{\cos }^2}A - {{\sin }^2}A} \right)\sin A}} .
Applying the identity a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)(a - b) , we get, (cos⁡A−sin⁡A)cos⁡2A(cos⁡A−sin⁡A)(cos⁡A+sin⁡A)sin⁡A\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, cos⁡2A(cos⁡A+sin⁡A)sin⁡A\dfrac{{{{\cos }^2}A}}{{\left( {\cos A + \sin A} \right)\sin A}} .
Now, consider right-hand side, cot⁡A1+tan⁡A\dfrac{{\cot A}}{{1 + \tan A}} , convert it into sin⁡A\sin A and cos⁡A\cos A form, we get, cos⁡Asin⁡A1+sin⁡Acos⁡A\dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{1 + \dfrac{{\sin A}}{{\cos A}}}} .
Taking least common multiple, and solving, cos⁡2Asin⁡A(cos⁡A+sin⁡A)\dfrac{{{{\cos }^2}A}}{{\sin A(\cos A + \sin A)}} , which is equal to the left-hand side.
Thus, cos⁡2Asin⁡A(cos⁡A+sin⁡A)=cos⁡2Asin⁡A(cos⁡A+sin⁡A)\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, cot⁡A−12−sec⁡2A\dfrac{{\cot A - 1}}{{2 - {{\sec }^2}A}} and convert each term into tan⁡A\tan A form, we get, 1tan⁡A−12−(1+tan⁡2A)\dfrac{{\dfrac{1}{{\tan A}} - 1}}{{2 - (1 + {{\tan }^2}A)}} , taking least common multiple and solving, we get, 1−tan⁡Atan⁡A1−tan⁡2A\dfrac{{\dfrac{{1 - \tan A}}{{\tan A}}}}{{1 - {{\tan }^2}A}} , which gives, 1−tan⁡Atan⁡A(1−tan⁡2A)\dfrac{{1 - \tan A}}{{\tan A(1 - {{\tan }^2}A)}} . Now, using the identity, a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)(a - b) , 1−tan⁡Atan⁡A(1−tan⁡A)(1+tan⁡A)\dfrac{{1 - \tan A}}{{\tan A(1 - \tan A)(1 + \tan A)}} which gives 1tan⁡A(1+tan⁡A)\dfrac{1}{{\tan A\left( {1 + \tan A} \right)}} which can also be written as cot⁡A(1+tan⁡A)\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\cos A and sin⁡A\sin A form and simplify.