Question

Prove that:
((cos x + sin x)/(cos x - sin x))
-((cos x - sin x)/(cos x + sin x))
= 2 tan 2x​

Answer 1

Answer:

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Answer 2

Answer:

$\bold{\underline{To \ Proof:}}$

$\sf \frac{cos \: x + sin \: x}{cos \: x - sin \: x} - \frac{cos \: x - sin \: x}{cos \: x + sin \: x} = 2tan \: 2x$

$\bold{\underline{Proof:}}$

LHS:

$\sf \implies \frac{cos \: x + sin \: x}{cos \: x - sin \: x} - \frac{cos \: x - sin \: x}{cos \: x + sin \: x} \\ \\ \sf \implies \frac{(cos \: x + sin \: x)(cos \: x + sin \: x)}{(cos \: x + sin \: x)(cos \: x - sin \: x)} - \frac{(cos \: x - sin \: x)(cos \: x - sin \: x)}{(cos \: x + sin \: x)(cos \: x - sin \: x)} \\ \\ \sf \implies \frac{(cos \: x + sin \: x)(cos \: x + sin \: x) - (cos \: x - sin \: x)(cos \: x - sin \: x)}{(cos \: x + sin \: x)(cos \: x - sin \: x)} \\ \\ \sf \implies \frac{ {(cos \: x + sin \: x)}^{2} - {(cos \: x - sin \: x)}^{2} }{ {cos}^{2}x - {sin}^{2} x} \\ \\ \sf \implies \frac{ {cos}^{2}x + {sin}^{2}x + 2cos \: x.sin \: x - {cos}^{2}x - {sin}^{2}x + 2cos \: x.sin \: x }{ {cos}^{2} x - {sin}^{2} x} \\ \\ \sf \implies \frac{4cos \: x.sin \: x}{{cos}^{2} x - {sin}^{2} x} \\ \\ \sf {sin}^{2} x = 1 - {cos}^{2} x : \\ \sf \implies \frac{4cos \: x.sin \: x}{{cos}^{2} x - (1 - {cos}^{2} x)} \\ \\ \sf \implies \frac{4cos \: x.sin \: x}{2{cos}^{2} x - 1} \\ \\ \sf 2cos \: x.sin \: x = sin \: 2x : \\ \sf \implies \frac{2sin \: 2x}{2{cos}^{2} x - 1} \\ \\ \sf 2{cos}^{2} x - 1 = cos \: 2x : \\ \sf \implies \frac{2sin \: 2x}{cos \: 2x} \\ \\ \sf \implies 2tan \: 2x$

RHS:

$\sf \implies 2tan \: 2x$

$\therefore$

$\bold{LHS = RHS}$

$\underline{ \sf Hence \: Proved}$