Math, asked by sumeramalik, 8 months ago

1+sin/1-sin=(sec+tan)²​

Answers

Answered by gauravjacker9602
0

Step-by-step explanation:

Can you prove the identity: (sec A - tan A) ^2=1-sin A/1+sin A?

Looking for the right CA guidance? Look no further.

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

Solution

To prove choose the complex side. In this equation both are equally complex. You can choose either side and methodically simplify moving towards the goal side. Apply the formulae correctly.

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

*******************************************

Alternatively,

LHS

= (sec A - tan A)²; express all as sine and cosine

= (1/cos A - sin A/cos A)²

= [(1 - sin A)/cos A)]²

= (1 - sin A)²/cos² A

= (1 - sin A)²/(1 - sin² A)

= (1 - sin

Continue Reading

How do I prove that (1 - sin A) / (1 + sin A) = (sec A - tan A)?

How do we prove 1+sinA/1-sinA= (secA+tanA) ^2?

If secA+tanA=p, then what is sin A?

How we can prove secA = (l^2+1) /2l, if sec A+tan A= l?

How do I prove this identity, “cosec(A) sec^2(A) =cosec(A) +tan(A) sec(A)”?

(secA−tanA)2=1−sinA1+sinA

Considering LHS,

(secA−tanA)2=(1cosA−sinAcosA)2

=(1−sinAcosA)2

=((1−sinA)2cos2A)

=((1−sinA)21−sin2A)

=((1−sinA)2(1−sinA)(1+sinA))

=1−sinA1+sinA=RHS

Similar questions