Question

Prove that :1 - {tan}^{4} A/1 + {tan}^{4} a=cosa+sinA/cosA-sinA


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Answer 1

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

Step-by-step explanation:

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

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

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

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

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

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

Using the algebraic identity,

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

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

Using the trigonometric identity,

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

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

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

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

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

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

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

= $\dfrac{1+\sin 2A}{\cos 2A}$  [ ∵ $\cos 2A=\cos^2 A-\sin^2 A$

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