Math, asked by pjayanth23, 8 months ago

tana-sina/sin^3A=secA/(1+cosa)​

Answers

Answered by praveen2007h
1

LS = (tanA-sinA)/sin^2 A

= (sinA/cosA - sinA)/sin^2 A

= ( (sinA - sinAcosA)/cosA) /sin^2 A

= (sinA(1 - cosA)/cosA)/sin^2 A

= (1 - cosA)/(sinAcosA)

RS = (sinA/cosA) / (1 + cosA)

= (sinA/cosA) / (1 + cosA) * (1-cosA)/(1-cosA)

= (sinA/cosA)(1-cosA)/(1 - cos^2 A)

= (sinA/cosA)(1-cosA)/sin^2 A

= (1-cosA)/(sinAcosA)

= LS

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