Question

guys solv a question ​

Answer 1

Answer:

$\frac{cot^{2} A (sec A - 1)}{1 + sin A} = \frac{sec^{2} A (1 - sin A) }{sec A + 1}$

Step-by-step explanation:

$LHS = \frac{cot^{2} A (sec A - 1)}{1 + sin A}$

$= \frac{cot^{2} A (sec A - 1)}{1 + sin A} \frac{sec A + 1}{sec A + 1}$

$= \frac{cot^{2} A (sec^{2} A - 1)}{(1 + sin A)(sec A + 1)}$

$= \frac{cot^{2} A (tan^{2} A)}{(1 + sin A)(sec A + 1)}$

$= \frac{1}{(1 + sin A)(sec A + 1)}$

$= \frac{1}{(1 + sin A)(sec A + 1)} \frac{1 - sin A}{1 - sin A}$

$= \frac{1 - sin A}{(1 - sin^{2} A (sec A + 1) )}$

$= \frac{1 - sin A}{cos^{2} A (sec A + 1) )}$

$= \frac{sec^{2} A (1 - sin A) }{sec A + 1}$

= RHS

Hence, LHS = RHS

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