Question

Prove that SinA/(1+CosA)=(CosecA-CotA)​

Answer 1

Please see the attachment!

Answer 2

$\bold{GIVEN}$

$\huge \boxed{\mathsf{\dfrac{sin \: A}{1+cos \: A} = cosec \: A - cot \: A}}$

On taking left handside,

$\mathsf{\dfrac{sin \: A}{1+cos \: A}}$

Multiply the numerator as well as denominator with

$\LARGE \boxed{\mathsf{1- cos \: A}}}$

$\mathsf{\dfrac{sin \: A \times 1- cos \: A }{1+cos \:A\times 1- cos\:A}}$

SINCE,

$\mathsf{(a+b)(a-b)= a^{2}+ b^{2} }$

so, we can say

$\mathsf{(1+Cos \: A)(1- Cos \:A)= 1 - cos^{2}A}$

So,

$\mathsf{\dfrac{sin \:A (1- cos\:A)}{1 - cos^{2}A}}$

$\mathsf{\dfrac{sin \:A - sin \: A . cos\:A}{1 - cos^{2}A}}$

SINCE,

$\boxed{\mathsf{1 - cos^{2}A= sin^{2}A}}$

$\mathsf{\dfrac{sin \:A - sin \: A . cos\:A}{sin^{2}A}}$

on taking each one sepeartely,

$\mathsf{\dfrac{sin \:A }{sin^{2}A} - \dfrac{cos \: A .sin \:A }{sin^{2}A}}$

on solving we get

$\mathsf{\dfrac{1 }{sin \: A} - \dfrac{cos\:A}{sin \: A}}$

$\boxed{\mathsf{\frac{1}{sin\: A}= cosec \: A}}$

$\boxed{\mathsf{\frac{cos \: A}{sin\: A}= cot \: A}}$

THEREFORE,

$\mathsf{cosec \: A -cot \: A}$

L.H.S = R.H.S

$\large \mathsf{HENCE \: PROVE}$