tanA - sin A /sin^2A= tan A / 1 + cosA

Please prove this

4 answers

Did you mean:

(tanA - sinA)/tan^2 A = tanA/(1+cosA)

the way you typed it, it is not true

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
Thanks reiny sir
Excellent
Excellent