Question

Prove this:)

Answer 1

hence proved

Answer 2

lolz.........

Answer 3

wait

Answer 4

k!

Answer 5

main kyan karun jab tak
Aaram haram hai? @mathslover

Answer 6

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

Answer 7

k!

Answer 8

\[\frac{1-tanx}{1+tanx} = \frac{1 - \frac{sinx}{cosx}}{1+ \frac{sinx}{cosx}} = \frac{cosx - sinx}{cosx+sinx}\]

\[\frac{cosx-sinx}{cosx+sinx} = \frac{(cosx-sinx)(cosx+sinx)}{(cosx+sinx)^2}\]\[\frac{(cosx-sinx)(cosx+sinx)}{(cosx+sinx)^2} =\frac{\cos^2x-\sin^2x}{\cos^2x +\sin^2x +2sinxcosx}\]\[=\frac{1-2\sin^2x}{1 +2sinxcosx}\]

Answer 9

thanx I gt it?

Answer 10

\[= \frac{1-2\sin^2x}{1 + \sin2x}\]