Question

proof (tan A + secA)^2 is equal to 1 + sin A/ 1 - sin A​

Answer 1

Good Morning

Here Is Your Answer:-

To prove

(tan A + sec A)^2 = 1 + sin A / 1 - sin A

Proof:-

LHS:- 1 + sin A / 1 - sin A

RHS:- (tan A + sec A)^2

Then ,

RHS:- (tan A + sec A)^2 {tan A= sinA/

cosA}

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

= ( 1+sin A/cos A )^2 {sec A = 1/cosA}

= (1 + sin A )^2 ÷( cos A )^2

= (1+sin A) (1+sin A)÷cos^2A

= (1+ sinA) (1+sinA)÷(1-sin^2A) {1-sin^2A=

a^2-b^2}

= ( 1+sinA) (1+sin A)÷ (1+sinA)(1-sinA)

= (1+sinA) / (1-sinA)=LHS

(proved)

So,It is proved that LHS=RHS

I Hope It will Help You

Answer 2

Answer:

RHS

RHS= (1-sinA)/(1+sinA)

RHS= (1-sinA)/(1+sinA)= (1-sinA)/(1+sinA) × (1-sinA)/(1-sinA) By Rationalization

RHS= (1-sinA)/(1+sinA)= (1-sinA)/(1+sinA) × (1-sinA)/(1-sinA) By Rationalization= (1-sinA)²/(1+sinA)(1-sinA)

RHS= (1-sinA)/(1+sinA)= (1-sinA)/(1+sinA) × (1-sinA)/(1-sinA) By Rationalization= (1-sinA)²/(1+sinA)(1-sinA)= (1-sinA)/(1+sinA) = LHS