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# Can you prove this identity?

((1-sint)^2/cos^2t) = (1-sinx/ 1+sinx)

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- 1 decade agoFavorite Answer
L.H.S.=(1-sin x)^2/cos^2x

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

=(1-sin x)(1-sin x)/(1+sin x)(1-sin x)

=(1-sin x)/(1+sin x)

=R.H.S. [Proved]

- Anonymous1 decade ago
cos^2t = 1 - sin^2t = (1-sint)(1+sint)

so,

( 1- sint)^2 /(1-sint)(1+sint) =(1-sint)/(1+sint)

- mohanrao dLv 71 decade ago
(1 - sin(t))^2/cos^2(t)

=>[1-sin(t)]^2/(1 - sin^2(t))

write [1- sin(t)]^2 = [1-sin(t)][1-sin(t)] and

1 - sin^2(t) = [1 + sin(t)][1 - sin(t)] (since a^2-b^2 = (a+b)(a-b))

=>[1-sin(t)][1 - sin(t)]/(1+sin(t))(1- sin(t))

cancel (1- sin(t)) from both numerator and denominator

=>[1-sin(t)]/[1+ sin(t)]

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