Question

Prove the following-
$\dfrac{{cotA - cosA}}{{cotA + cosA}} = \dfrac{{cosecA - 1}}{{cosecA + 1}}$

Answer 1

Hint: We need to convert the trigonometric functions in this question. In such types of questions, the best method is to convert all the functions in terms of sine and cosine and then simplify to prove the two expressions equal. Some functions to be used are-
$\begin{aligned} &cotA; = \dfrac{{cosA}}{{sinA}}...\left( 1 \right) \\\ &cosecA; = \dfrac{1}{{sinA}}...\left( 2 \right) \\\ \end{aligned}$

Complete step-by-step answer:
We need to prove that-
$\dfrac{{cotA - cosA}}{{cotA + cosA}} = \dfrac{{cosecA - 1}}{{cosecA + 1}}$
So, we will proceed using property (1) as-
$\begin{aligned} &= \dfrac{{cotA - cosA}}{{cotA + cosA}} \\\ &= \dfrac{{\dfrac{{cosA}}{{sinA}} - cosA}}{{\dfrac{{cosA}}{{sinA}} + cosA}} \\\ \end{aligned}$
Taking cosA common,
= \dfrac{{cosA\left( {\dfrac{1}{{sinA}} - 1} \right)}}{{cosA\left( {\dfrac{1}{{sinA}} + 1} \right)}} \\\
Using property $\left( 2 \right)$, we can write that,
= \dfrac{{cosA\left( {cosecA - 1} \right)}}{{cosA\left( {cosecA + 1} \right)}} = \dfrac{{cosecA - 1}}{{cosecA + 1}} \\\

Hence, the given equation is proved.

Note: Even if we are unable to prove the two expressions equal directly, we can leave the first expression in a simplified form in terms of sine and cosine functions, and then convert the second expression into sine and cosine and then prove those values equal to each other.