Question

SinA/1-CosA + TanA/1+CosA=Cos²A+1/SinA×CosA

Answer 1

Correct Question :-

Prove that : sinA/(1 - cosA) + tanA/(1 + cosA) = (cos²A + 1)/sinA * cos A

Solution :-

$\sf \dfrac{sinA}{1 - cosA} + \dfrac{tanA}{1 + cosA} = \dfrac{cos^{2} A + 1}{sinA \times cosA}$

Consider RHS

$\sf = \dfrac{sinA}{1 - cosA} + \dfrac{tanA}{1 + cosA}$

Taking LCM

$\sf = \dfrac{sinA(1 + cosA) + tanA(1 - cosA}{(1 - cosA)(1 + cosA)}$

$\sf = \dfrac{sinA+ sinA.cosA + tanA - tan A.cosA}{ {1}^{2} - cos ^{2} A}$

[ Because (x - y)(x + y) = x² - y² ]

$\sf = \dfrac{sinA+ sinA.cosA + tanA - \dfrac{sinA}{cosA} .cosA}{1 - cos ^{2} A}$

[ Because tanA = sinA/cosA ]

$\sf = \dfrac{sinA+ sinA.cosA + tanA - sinA}{sin^{2} A}$

$\sf = \dfrac{sinA.cosA + tanA}{sin^{2} A}$

$\sf = \dfrac{sinA.cosA + \dfrac{sinA}{cosA} }{sin^{2} A}$

Taking sinA common in numerator

$\sf = \dfrac{sinA \bigg(cosA + \dfrac{1}{cosA} \bigg)}{sin^{2} A}$

$\sf = \dfrac{ cosA + \dfrac{1}{cosA}}{sin A}$

Taking LCM in numerator

$\sf = \dfrac{ \dfrac{cos^{2} A +1 }{cosA}}{sin A}$

$\sf = \dfrac{cos^{2} A +1 }{sinA.cosA}$

Hence proved.