SecA - 1/ SecA + 1 = Sin^2A /( 1 + CosA)^2
please prove it fastly
Arya79:
it's an math's question not of social science
Answers
Answered by
5
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
0
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
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