Question

Prove that
\[{1-\tan^2x \over 1+ \tan^2x }= 1-2\sin^2x\]

Answer 1

LHS:
\[(\cos^2x - \sin^2x)/(\cos^2x + \sin^2x)\]
numerator is Cos2x and denominator is 1
now Cos2x is\[1 - 2\sin^2x\]
= RHS
henc proved...

Answer 2

I don't understand. Where did the identities come from?

Answer 3

put tanx = Sinx/Cosx in first step.....

Answer 4

\(\large 1 + \tan^2 x = \sec ^2 x\) so ....

\(\LARGE \frac{1 - \tan^2 x}{\sec^2 x} = \frac{1}{\sec^2 x} - \frac{\tan^2 x}{\sec^2 x} = \cos^2 x - \sin^2 x\)

= \(\large 1 - \sin^2 x - \sin^2 x = 1 - 2\sin ^2 x = RHS\)

Answer 5

now, do u knw that cos(A+B) = cosAcosB - SinASinb

Answer 6

alternate solution :D

Answer 7

i'll let @Savvy explain his solution..carry on ^_^

Answer 8

@order do you understand my solution? :)

Answer 9

Igbasallote, I understand yours, but not why you got -tan^2x/sec^2x to be - sin^2x

Answer 10

put tanx = sinx/cosx and secx = 1/cosx

Answer 11

sorry for that..

\(\LARGE \tan ^2 x = \frac{\sin^2 x}{\cos ^2 x}\)

\(\LARGE \sec^2 x = \frac{1}{\cos^2 x}\)

\(\LARGE \frac{\tan ^2 x}{\sec^2 x} = \frac{(\frac{\sin ^2 x}{\cancel{\cos^2 x}})}{(\frac{1}{\cancel{\cos^2 x}})}\)

Answer 12

Thank you!! :D