prove that tanA+sinA by tanA-sinA=secA+1 by sec+1
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Answer:
sin
tanA+sinA/tanA-sinA=sinA/cosA+sinA/sinA/cosA-sinA
=sinA+sinAcosA/cosA/sinA-sinAcosA/cosA
so cosA cancel
= sinA+sinAcosA/sinA-sinAcosA
in this we take sinA common in numerator and denominator
=sinA(1+cosA/sinA)/sinA(1-cosA/sinA)
sinA cancel
=1+cosA/1-cosA.
cosA=1/secA
=1+1/secA/1-1/secA
=secA+1/secA/secA-1/SecA
so cancel secA
=secA+1/secA-1
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