Question

establish/prove this identity

Answer 1

Note : In the solution given below, theta is replaced by angle A.

Answer:

$\dfrac{sinA+1}{cosA}$

Step-by-step explanation:

= > ( sinA - cosA + 1 ) / ( sinA + cosA - 1 )

= > { sinA + 1 + cosA } / { sinA - 1 + cosA }

= > { √( sinA + 1 )^2 + √( 1 - sin^2 A ) } / { √( sinA - 1 )^2 + √( 1 - sin^2 A ) } { sinA + 1 = √( sinA + 1 )^2 & cos^2 A = 1 - sin^2 A }

$\implies \dfrac{\sqrt{(sinA+1)(sinA+1)}+\sqrt{(1-sinA)(1+sinA)}}{\sqrt{(sinA-1)(sinA-1)}+\sqrt{(1-sinA)(1+sinA)}}\\\\\\\implies \dfrac{\sqrt{(sinA+1)}\{\sqrt{(sinA+1)} + \sqrt{(sinA-1)}\}}{\sqrt{(sinA-1)}\{\sqrt{(sinA-1)}+\sqrt{(sinA+1)}\}}\\\\\\\implies\dfrac{\sqrt{sinA+1}}{\sqrt{sinA-1}}\\\\\\\implies\sqrt{\dfrac{sinA+1}{sinA-1}\times\dfrac{sinA+1}{sinA-1}}\\\\\\\implies\sqrt{\dfrac{(sinA+1)^2}{sin^2-1}}\\\\\\\implies\dfrac{sinA+1}{cosA}$

Hence proved.