Asked by Venkatesh
tanA - sin A /sin^2A= tan A / 1 + cosA
Please prove this
Please prove this
Answers
Answered by
Reiny
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
(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
Answered by
Venkatesh
Thanks reiny sir
Answered by
Salma
Excellent
Answered by
Wow
Excellent
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