Question: Prove that: $\dfrac{cos2A}{1+ sin2A} = \dfrac{1- tanA}{1+ tanA}$
Solution:
Taking LHS,
$= \dfrac{cos2A}{1+ sin2A}$
Expanding cos2A and sin2A and identity of 1
$= \dfrac{cos^2A - sin^2A}{sin^2A + cos^2A + sin2A}$
Performing simple mathematical operations
$= \dfrac{(cosA + sinA)(cosA - sinA)}{(cosA + sinA)^2}$
$= \dfrac{cosA - sinA}{cosA + sinA}$
Dividing each term by cosA
$= \dfrac{\frac{cosA}{cosA} - \frac{sinA}{cosA} }{\frac{cosA}{cosA} + \frac{sinA}{cosA}}$
We know, [$\frac{sinA}{cosA} = tanA$]
$= \dfrac{1 - tanA}{1 + tanA}$
= RHS
#proved
4 Comments
how did (cosA+sinA)whole square become????? in step no 3
ReplyDeletesin2A = 2sinAcosA
Deleteand,
$cos^2A + sin^2A + 2sinAcosA$
= $(cosA +sinA)^2$
$= \dfrac{\frac{cosA}{cosA} - \frac{sinA}{cosA} }{\frac{cosA}{cosA} + \frac{sinA}{cosA}}¢
ReplyDeletewhat is this
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