Question

(1-tan^4x)/(1+tan^4x)=(Cosx+sinx)/(Cosx-sinx) prove it plzz my exams are tomorrow.

For 100 points and brainlist answer.​

Answer 1

Step-by-step explanation:

manipulate the left side

⇒

1

+

tan

x

1

−

tan

x

=

1

+

sin

x

cos

x

1

−

sin

x

cos

x

rewrite numerator / denominator as single fractions.

=

cos

x

+

sin

x

cos

x

cos

x

−

sin

x

cos

x

multiply the fractions by ' inverting' the fraction on denominator.

⇒

cos

x

+

sin

x

cos

x

×

cos

x

cos

x

−

sin

x

=

cos

x

+

sin

x

cos

x

−

sin

x

=

right side

Answer 2

$\dfrac{1-\tan^4 x}{1+\tan^4 x} =\dfrac{\cos x+\sin x}{\cos x-\sin x}$, proved.

Step-by-step explanation:

Prove that, $\dfrac{1-\tan^4 x}{1+\tan^4 x} =\dfrac{\cos x+\sin x}{\cos x-\sin x}$.

L.H.S.$=\dfrac{1-\tan^4 x}{1+\tan^4 x}$

$=\dfrac{1-\dfrac{\sin^4 x}{\cos^4 x}}{1+\dfrac{\sin^4 x}{\cos^4 x}}$

$= \dfrac{\cos^4 x-\sin^4 x}{\cos^4 x+\sin^4 x}$

$=\dfrac{(\cos^2 x)^2-(\sin^2 x)^2}{(\cos^2 x)^2+(\sin^2 x)^2}$

$= \dfrac{(\cos^2 x+\sin^2 x)(\cos^2 x-\sin^2 x)}{(\cos^2 x)^2+(\sin^2 x)^2}$

Using the algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

$= \dfrac{(1)(\cos^2 x-\sin^2 x)}{(\cos^2 x+\sin^2 x)^2-2\cos^2 x\sin^2 x}$

Using the trigonometric identity,

$\sin^2 \theta+\cos^2 \theta = 1$

$=\dfrac{(\cos x+\sin x)(\cos x-\sin x)}{(1)^2-2\cos^2 x.\sin^2 x}$

$= \dfrac{1+\sin 2x}{\cos 2x}$

R.H.S. = $\dfrac{\cos x+\sin x}{\cos x-\sin x}$

$= \dfrac{\cos x+\sin x}{\cos x-\sin x}\times \dfrac{\cos x+\sin x}{\cos x+\sin x}$

$= \dfrac{(\cos x+\sin x)^2}{\cos^2 x-\sin^2 x}$

$=\dfrac{\cos^2 x+\sin^2 x+2\sin x\cos x}{\cos^2 x-\sin^2 x}$

$= \dfrac{1+\sin 2x}{\cos 2x}$

∴ L.H.S. = R.H.S. $= \dfrac{1+\sin 2x}{\cos 2x}$

Hence, $\dfrac{1-\tan^4 x}{1+\tan^4 x} =\dfrac{\cos x+\sin x}{\cos x-\sin x}$, proved.