Question

1-tan^2(theta)/1+tan^2(theta)=1-2sin^2(theta)
Prove identity

Answer 1

Do you know a common Pythagorean identity associated with this?

Answer 2

Sin^2(theta)+Cos^2(theta)=1

Answer 3

tan^2(theta)=Sin^2(theta)/cos^2(theta)

Answer 4

well, you can actually just use the second thing. forget identities.

Answer 5

i used it and still cant reach to 1-2sin^2(theta)

Answer 6

what are you getting?

Answer 7

1-(Sin^2(theta)/cos^2(theta))/1+(sin^2(theta)/Cos^2(theta)

Answer 8

Get a common denominator in both the numerator and denominator.

Answer 9

Then cancel that denominator.

Answer 10

How to get a common denominator and they are already similar but the sign are different

Answer 11

@ParthKohli it would be easy if u simplify both R.H.S and L.H.S and go to cos 2x .. R.H.S is already in the expanded form of cos 2x and under 3 steps u could reach cos^x - sin^x from L.H.S which is also cos 2x :)

Answer 12

\[1 - \tan^2\theta = 1 - \dfrac{\sin^2\theta}{\cos^2 \theta} = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta}\]

Answer 13

similarly in the denominator, you have\[\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta}\]Cancel the \(\cos^2\theta\)

Answer 14

also, \(\sin^2\theta + \cos^2 \theta = 1\).

Answer 15

So\[\dfrac{\cos^2\theta- \sin^2\theta}{\sin^2 \theta + \cos^2\theta} = \cos^2\theta - \sin^2\theta\]

Answer 16

now in \(\cos^2\theta - \sin^2\theta\), substitute \(1 - \sin^2\theta\) for \(\cos^2\theta\)

Answer 17

Thank you for your help :)

Answer 18

You're welcome!