prove that secA-1/secA+1=(sinA/1+cosA)2
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Answer:
Step-by-step explanation:
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
Answered by
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Answer:
answer for the given problem is given
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