if sin tita + cos tita=p and sec tita + cosec tita =q then prove that q(p^2-1)=2p
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Step-by-step explanation:
L.H.S q(p²-1)= [(sin tita + cos tita)²-1] (sec tita + cosec tita) = 2sin tita cos tita (1/cos tita + 1/cosec tita) =2(sin tita + cos tita)
R. H. S 2p= 2(sin tita + cos tita)
that is, L. H. S= R. H. S (proved)
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Answer:
Taking L.H.S
=q(p^2-1)
=(secA+cosecA){(sinA+cosA)^2-1}
=(secA+cosecA){sin^2A+cos^2A+2sinA.cosA-1}
=(secA+cosecA)(1+2sinA.cosA-1) (USING
sin^2A+cos^2A=1)
=(secA+cosecA)(2sinA.cosA)
=(1/cosA+1/sinA)(2sinA.cosA)
Taking L.CM
=(cosA+sinA)(2sinA.cosA)/sinA.cosA
=(cosA+sinA)(2)
=2(sinA+cosA)
=2p (given that sinA+cosA=p)
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