Question

prove that sinA/1+cosA+sinA/1-cosA=2secA

Answer 1

Answer:

Step-by-step explanation:

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Coz RHS comes out to be 2cosecA

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Answer 2

Answer:

on solving the left hand side of the expression:

$\dfrac{\sin A}{1+\cos A}+\dfrac{\sin A}{1-\cos A}$

on taking the l.c.m of the above expression we get:

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

since on cancelling the same term with opposite signs in the numerator.

$=\dfrac{2\sin A}{\sin^2 A}\\\\=\dfrac{2}{\sin A}\\\\=2 cosec A$

Hence, the equality is written as:

$\dfrac{\sin A}{1+\cos A}+\dfrac{\sin A}{1-\cos A}=2 cosec A$