Question

Prove it..............​

Answer 1

Answer:

LHS

sin a/(1+sina)

=sina(1-sin a)/(1+sin a)(1-sin a)

=(sin a - sin^2a)/(1^2-sin ^2 a)

=(sin a- sin^2a)/(1-sin^2a)

=(sin a- sin ^2 a)/cos^2a

=(sina/cos^2a)-(sin^2a/cos^2a)

=(sina/cos a)×(1/cos a)-tan^2a

=tana×sec a -tan^2a

Step-by-step explanation:

hope it helps you

mark brainliest

Answer 2

Step-by-step explanation:

Hey mate☺️

Here's your Answer

We can convert RHS to LHS to Prove that!!

SecA=1/cosA

tanA=sinA/CosA

tan²A=sin²A/cos²A

therefore

RHS

[(1/cosA)*SinA/CosA]-(Sin²A/Cos²A)

=[sinA/Cos²A]-(Sin²A/Cos²A)

= sinA-sin²A

Cos²A

= sinA(1-SinA)

1-sin²A

= sinA(1-SinA)

(1+sinA)(1-sinA)

= sinA.

(1+SinA)

So LHS=RHS

Hence Proved

Thank you

Regards☺️