Question

Prove this identity: $\frac{1+cos\alpha }{sin\alpha ^{2} } = \frac{1}{1-cos\alpha }$ I had asked this question before but it had a typo so can you please solve

Answer 1

Question

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Prove this identity :

$\sf{\frac{1+cos\alpha}{sin^2\alpha}=\frac{1}{1-cos\alpha}}$

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Solution

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Taking LHS

$:\implies\bf{LHS}=\sf{\frac{1+cos\alpha}{sin^2\alpha}}$

Multiplying by $\sf{\frac{1-cos\alpha}{1-cos\alpha}}$

$:\implies\bf{LHS}=\sf{\frac{1+cos\alpha}{sin^2\alpha}\times\frac{1-cos\alpha}{1-cos\alpha}}$

Using identity (a+b)(a-b)=a²-b²

$:\implies\bf{LHS}=\sf{\frac{(1)^2-(cos\alpha)^2}{sin^2\alpha(1-cos\alpha)}}\\\\:\implies\bf{LHS}=\sf{\frac{1-cos^2\alpha}{sin^2\alpha(1-cos\alpha)}}$

Using trigonometric identity

1 - cos²A = sin²A

$:\implies\sf{LHS=\frac{sin^2\alpha}{sin^2\alpha(1-cos\alpha)}}\\\\:\implies\sf{LHS=\frac{1}{1-cos\alpha}=RHS}$

Hence Proved .

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More trigonometric identities  

→ sin²A + cos²A = 1

→ 1 + tan²A = sec²A

→ 1 + cot²A = cosec²A

→ sin(90-A) = cos A

→ cos(90-A) = sin A

→ sec(90-A) = cosec A

→ cosec(90-A) = sec A

→ tan(90-A) = cot A

→ cot(90-A) = tan A

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Answer 2

${ \huge{ \underline{ \underline{ \red{ \mathfrak{Question:-}}}}}}$



▪ Prove the identity




${ \bold{ \huge{ \blue{ \frac{1 + \cos( \alpha ) }{ { \sin( \alpha ) }^{2} } = \frac{1}{1 - \cos( \alpha ) } }}}}$




${ \huge{ \underline{ \underline{ \red{ \mathfrak{Solution:-}}}}}}$




${\huge{ \bold{ \frac{1 + \cos( \alpha ) }{ { \sin( \alpha ) }^{2} } = \frac{1}{1 - \cos( \alpha ) }} }}$



▪ solving the L.H.S......



${\huge { \bold{ = \frac{1 + \cos( \alpha ) }{ { \sin( \alpha ) }^{2} }} }}$



▪ using the trigonometric identity.....



${ \boxed{ \red{ \bold{ \: \: { \sin( \alpha ) }^{2} = 1 - { \cos( \alpha ) }^{2} \: \: }}}}$



${ \huge {\bold{ = \frac{1 + \cos( \alpha ) }{1 - { \cos( \alpha ) }^{2} }} }}$




▪ we know that.....




${ \boxed{ \bold{ \red{ \: {a}^{2} - {b}^{2} = (a - b)(a + b) \: }}}}$



▪ using this algebraic identity in the denominator.....



${ \huge {\bold{ = \frac{(1 + \cos( \alpha ) )}{(1 - \cos( \alpha )) (1 + \cos( \alpha ) )} }}}$



${\huge{ \bold{ = \frac{1}{1 - \cos( \alpha ) }} }}$



thus,



${ \boxed{ \bold{ \blue{ \: L.H.S. = R.H.S. \: \: }}}}$



hence, proved....



${ \boxed{ \bold{ \red{ \frac{1 + \cos( \alpha ) }{ { \sin( \alpha ) }^{2} } = \frac{1}{1 - \cos( \alpha ) } }}}}$






.