Question

root 1+sinA/1-sinA = secA +tanA​

Answer 1

Answer:

Step-by-step explanation:

√ (1 + SinA/1 - SinA)

//multiply numerator and denominator by 1 - SinA

=> √ [(1 + SinA/1 - SinA) * (1 + SinA/1 - SinA)

=> √ [ (1 + SinA)²/ (1 - SinA)(1 + SinA)]

=> √ [ (1 + SinA)²/ (1 - Sin²A)]  (∵ In denominator (a + b)(a-b) = a² - b²)

=> √ [ (1 + SinA)²/Cos²A] (∵ 1 - Sin²A = Cos²A)

=> 1 + SinA / CosA

=> 1/CosA + SinA/CosA

=> SecA + TanA

= R.H.S

Hence proved.