Question

1-tan⁴theta by 1+tan⁴theta =cos theta+sin theta by cos theta - sin theta

Answer 1

Answer:

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Answer 2

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

Step-by-step explanation:

Prove that,

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

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

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

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

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

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

Using the identity,

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

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

Using the trigonometric identity,

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

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

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

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

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

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

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

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

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