Asked by kumar
sinA+cosA=p and secA+CosecA=q
then show that q[p^2-1]=2p
then show that q[p^2-1]=2p
Answers
Answered by
Arora
q[p^2-1]
= (secA + cosecA)[(sinA+cosA)^2 - 1]
= (secA + cosecA)[ (sin^2A + cos^2A) + 2sinAcosA - 1]
= (secA + cosecA)[ 1 + 2sinAcosA - 1 ]
= (secA + cosecA)(2sinAcosA)
= 2sinA + 2cosA
= 2(sinA + cosA) = 2p
= (secA + cosecA)[(sinA+cosA)^2 - 1]
= (secA + cosecA)[ (sin^2A + cos^2A) + 2sinAcosA - 1]
= (secA + cosecA)[ 1 + 2sinAcosA - 1 ]
= (secA + cosecA)(2sinAcosA)
= 2sinA + 2cosA
= 2(sinA + cosA) = 2p
Answered by
Nishu
Very good answer.THANKS
Answered by
Anonymous
Thanks
Answered by
nope
show the steps in a better way
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