Question

Prove that sin A + cos A / sinA − cos A + sinA − cos A / sin A + cos A = 2 / sin²A − cos²A .​

Answer 1

L.H.S = {[ sinA + cosA] / [sinA-cosA]} + {[sinA -cosA]/ [sinA+cosA]}

= {(sinA+cosA)^2 + ( sinA-cosA)^2}/{(sinA+cosA)(sinA-cosA)}

= (1+2sinAcosA +1 - 2sinAcosA) / (sin^2A - cos^2A)

= 2/(sin^2A - cos^2A)

= R.H.S

Hence proved

Answer 2

Step-by-step explanation:

$\tt We\:have,$

$\tt \dfrac{\sin θ + \cos θ}{\sin θ- \cos θ} +\dfrac{\sin θ - \cos θ}{\sin θ+\cos θ}$

$\tt \dfrac{(\sin θ + \cos θ)^2+(\sin θ- \cos θ)^2 }{(\sin θ- \cos θ)(\sin θ+ \cos θ)}$

$\dfrac{\sin^2 θ + \cos^2 θ+2\sin θ \cos θ+\sin ^2θ+ \cos θ^2-2\cos \theta \sin \theta}{ \sin ^2 θ - \cos ^2 θ}$

$\dfrac{1+2\sin θ \cos θ+1-2\cos \theta \sin \theta}{ \sin ^2 θ - \cos ^2 θ}$

$\dfrac{1+\cancel{2\sin θ \cos θ}+1-\cancel{2\cos \theta \sin \theta}}{ \sin ^2 θ - \cos ^2 θ}$

$\dfrac{2}{ \sin ^2 θ - \cos ^2 θ}$