Question

1+sin2θ / 1-sin2θ = tan^2 ( π/4 + θ)

Answer 1

So prove it?

Answer 2

yeah

Answer 3

Here use;

\[\sin(2 \theta) = 2\sin(\theta)\cos(\theta)\]

\[1 = \sin^2 ( \theta) + \cos^2 (\theta)\]

Answer 4

\[\frac{1 + \sin(2 \theta)}{1 - \sin(2 \theta)} = \frac{\sin^2 (\theta) + \cos^2 (\theta) + 2\sin(\theta)\cos(\theta)}{\sin^2 (\theta) + \cos^2 (\theta) - 2\sin(\theta)\cos(\theta)} \implies \frac{(\sin(\theta) + \cos(\theta))^2}{(\sin(\theta) - \cos(\theta))^2}\]

Answer 5

\[(\frac{\sin(\theta) + \cos(\theta)}{\sin(\theta) - \cos(\theta)})^2\]

Answer 6

okay .. then ?

Answer 7

Now divide by cos(theta) in the numerator and denominator:

\[(\frac{1 +\tan(\theta)}{1 - \tan(\theta)})^2 \implies (\frac{\tan(45) + \tan(\theta)}{1 - \tan(45) \tan(\theta)})^2 \implies \tan^2(45 + \theta)\]

Answer 8

\[45^{\circ} = \frac{\pi}{4}\]

Answer 9

\[\tan(45) = 1\]

Answer 10

(1+tan(θ)1−tan(θ))^2 how did yu get this ?

Answer 11

\[\large (\frac{\frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\cos(\theta)}}{\frac{\sin(\theta)}{\cos(\theta)} - \frac{\cos(\theta)}{\cos(\theta)}})^2 \implies (\frac{\tan(\theta) + 1}{\tan(\theta) -1 })^2\]

Answer 12

In square you can change the order:

so you can reverse the denominator:
\[(\frac{\tan(\theta) + 1}{1 - \tan(\theta) })^2\]

Answer 13

okay thank yu soo mch .. wish i had more medals for yu :D

Answer 14

It is about knowledge and not medals..

Welcome dear..

Answer 15

:)