seca-1/seca+1 =[sina / 1+cosa]^2
sidd47:
pleas help me
Answers
Answered by
41
hope it helps........
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mult no by ?? and ??
How do you get the step 4 answer
mutilplying first by seca-1 and then by 1+cosa
Answered by
17
note : mistake in RHS
Solution:
Given ,
LHS = (secA-1)/(secA+1)
= [(1/cosA)-1]/[(1/cosA)+1]
= [(1-cosA)/cosA]/[(1+cosA)/cosA]
after cancellation, we get
= (1-cosA)/(1+cosA)
multiply numerator and denominator by (1-cosA), we get
=[(1-cosA)(1-cosA)]/[(1+cosA)(1-cosA)]
= (1-cosA)²/[1²-cos²A]
= (1-cosA)²/sin²A
= RHS
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