Question

pls solve it

you have to prove the identity
R.H.S. =L.H.S.
correct answer will be brainlist ​

Answer 1

Solution

In above attachment answer is attached

Take LHS

$\implies$ (1 + tan A + cot A)(sin A - cos A)

$\implies$ sin A - cos A + tan A.sin A + cot A.sin A - tan A.cos A - cot A. cos A

$\implies sin\: A \: - \: cos\: A\: +\dfrac{sin\: A}{cos\: A}. sin \: A \: +\: \dfrac{cos\: A}{sin\: A}. sin\:A\: - \:\dfrac{sin\:A}{cos\:A}. cos\:A \: - \:\dfrac{cos\: A}{sin\:A}. cos\: A$

Eliminate same value

$\implies sin\: A \:-\: cos \: A \:+\:\dfrac{sin^{2}\:A}{cos\:A}\: + \:cos \: A \: - \:sin\:A\:-\:\dfrac{cos^{2}\:A}{sin\:A}$

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

$\implies sin\:A.tan\:A\:-\: cot\:A.cos\:A$

$\boxed{L.H.S\: =\: R.H.S}$

Hence proved