Question

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Answer 1

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❚ QuEstiOn ❚

prove that ,

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

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❚ ANsWeR ❚

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

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

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

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

$=\sf{\ \ {\dfrac{2}{\sin^2 A+\sin^2 A-1}}}$

$=\sf{\ \ {\dfrac{2}{2\sin^2 A-1}}}$

$=\sf{\ \ {R.H.S.= (proved)}}$

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❏ Used Formula :-

$\longrightarrow\sf{\ \ {(A+B)^2+(A-B)^2=2(A^2+B^2)}}$

$\longrightarrow\sf{\ \ {(A+B)^2-(A-B)^2=4AB}}$

$\longrightarrow\sf{\ \ {(A+B)(A-B)^2=A^2-B^2}}$

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Answer 2

Solution →

$\frac{ \sin(a) + \cos(a) }{ \sin(a ) - \cos(a) } + \frac{ \sin(a) - \cos(a) }{ \sin(a) + \cos(a) } = \frac{2}{2 {sin}^{2} a - 1 }$

Left Hand Side →

$= \frac{ \sin(a) + \cos(a) }{ \sin(a ) - \cos(a) } + \frac{ \sin(a) - \cos(a) }{ \sin(a) + \cos(a) }$

$= \frac{{( \sin(a) + \cos(a)) }^{2} + {( \sin(a) - \cos(a) ) }^{2} }{( \sin(a) - \cos(a) )( \sin(a) + \cos(a) )}$

We know the formulas →

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• (a-b)(a+b) = a² - b²

$</strong><strong>=</strong><strong>\frac{ { \sin}^{2} a + {cos}^{2} a + 2sin \: a \times cos \: a \: + \: {sin}^{2} a + {cos}^{2}a - 2sin \: a \times cos \: a }{ {sin}^{2}a - {cos}^{2} a }$

we know that →

• cos²a = 1- sin²a

$= \frac{2 {sin}^{2}a \: + \: 2 {cos}^{2} a }{ {sin}^{2}a - (1 - {sin}^{2} a) }$

$= \frac{2( {sin}^{2} a + {cos}^{2}a) }{ {sin}^{2}a - 1 + {sin}^{2}a }$

We know that →

• sin²a + cos²a = 1

$\boxed{ = \frac{2 }{2 \: {sin}^{2}a - 1 }}$ RHS

Hence proved !!

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