Question

✓1+sinA/1-sinA = secA+tanA​

Answer 1

Step-by-step explanation:

we have to prove,,

√{(1+sinA)/(1-sinA)} = secA+tanA

LHS,

by rationalizing,,

√{(1+sinA)²/(1-sin²A)}

√{(1+sinA)²/cos²A}

(1+sinA)/cosA

(1/cosA+sinA/cosA)

=> secA+tanA = RHS

hence proved....

Answer 2

Answer:

√1+sin A/1-sin A

Multiplying both numerator and denominator by 1+sin A , we get

= √(1+sin A)²/(1-sin A)(1+sin A)

= √(1+sin A)²/1-sin²A

= √(1+sin A)²/cos²A                { sin²Θ + cos²Θ = 1 }

=  1+sin A/ cos A                     { √a² = a }

=  1/cos A + sin A/cos A

=  sec A + tan A

=  RHS

Hence proved.