Question

SecA - 1/ SecA + 1 = Sin^2A /( 1 + CosA)^2

please prove it fastly​

Answer 1

L.H.S

(secA – 1)/ (sec A + 1)

=(1/cosA - 1)/ (1/cos A +1)

=(1 – cosA/cosA)/ (1 + cosA/cosA)

=(1 – cosA)/ (1 + cosA)

Multiplying numerator and denominator by (1 + cosA), we get

=(1 - cosA)(1 + cosA)/ (1 + cosA)2

=(1 – cos2A)/ (1 + cosA)2

=Sin2A/ (1 + cosA)2

=[sinA/(1 + cosA)]2

=sin^2A/(1 + cosA)^2 = R.H.S

Answer 2

Answer:

Explanation:

(secA – 1)/ (sec A + 1)

=(1/cosA - 1)/ (1/cos A +1)

=(1 – cosA/cosA)/ (1 + cosA/cosA)

=(1 – cosA)/ (1 + cosA)

Multiplying numerator and denominator by (1 + cosA), we get

=(1 - cosA)(1 + cosA)/ (1 + cosA)2

=(1 – cos2A)/ (1 + cosA)2

=Sin2A/ (1 + cosA)2

=[sinA/(1 + cosA)]2

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

PROVED