Math, asked by balirampatel4044, 11 months ago

Prove that
COS X + sin x
COS X - sin x
COS X - sin x
COS X + sin x
= 2 tan 2x.​

Answers

Answered by kush193874
10

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}

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