Question

Brainlians Here's your question
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Prove that.

$\sf\sqrt{\dfrac {1-cos(\alpha)}{1+cos(\alpha)}}=\dfrac {sin(\alpha)}{1+cos (\alpha)}$
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Answer 1

EXPLANATION.

$\sf \implies \sqrt{\dfrac{1 - cos(\alpha )}{1 + cos(\alpha )} }$

As we know that,

In this type of question we multiply and divide by 1 + cosα, we get.

$\sf \implies \sqrt{\dfrac{1 - cos(\alpha )}{1 + cos(\alpha )} \times \dfrac{1 + cos(\alpha )}{1 + cos(\alpha )} }$

$\sf \implies \sqrt{\dfrac{1 - cos^{2}(\alpha) }{(1 + cos(\alpha ))^{2} } }$

$\sf \implies \sqrt{\dfrac{ sin^{2}(\alpha) }{(1 + cos(\alpha ))^{2} } }$

$\sf \implies \dfrac{sin(\alpha )}{1 + cos(\alpha )}$

$\sf \implies \sqrt{\dfrac{1 - cos(\alpha )}{1 + cos(\alpha )} } = \dfrac{sin(\alpha )}{1 + cos(\alpha )}$

                                                                                                                         

MORE INFORMATION.

Fundamental trigonometric identities.

(1) = sin²x + cos²x = 1.

(2) = 1 + tan²x = sec²x.

(3) = 1 + cot²x = cosec²x.

Trigonometric ratio of multiple angles.

(1) = sin2x = 2sinx.cosx = 2tanx/1 + tan²x.

(2) = cos2x = cos²x - sin²x = 2cos²x - 1 = 1 - 2sin²x = 1 - tan²x/1 + tan²x.

(3) = tan2x = 2tanx/1 - tan²x.

(4) = sin3x = 3sinx - 4sin³x.

(5) = cos3x = 4cos³x - 3cosx.

(6) = tan3x = 3tanx - tan³x/1 - 3tan²x.

Answer 2

Answer:

To Prove :-

$\sf\sqrt{\dfrac {1-cos(\alpha)}{1+cos(\alpha)}}=\dfrac {sin(\alpha)}{1+cos (\alpha)}$

Required Proof :-

$\sqrt \dfrac{1 - \cos \alpha}{1 + \cos \alpha } = \sqrt{ \dfrac{ \sin \alpha }{1 + \cos \alpha} }$

Solving the LHS

$\sqrt{ \dfrac{1 - \cos \alpha }{1 + \cos \alpha } }$

$\sqrt{ \dfrac{ \bigg \lgroup1 - \cos \alpha \bigg \rgroup \times \bigg \lgroup1 + \cos \alpha \bigg \rgroup}{ \bigg \lgroup1 + \cos \alpha \bigg \rgroup \times \bigg \lgroup1 + \cos \alpha \bigg \rgroup}}$

Apply Identity

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

$\bf \red{(a + b)(a + b) = (a + b {)}^{2} }$

$\sf \: \sqrt{ \dfrac{1 - { \cos }^{2} \alpha }{ \bigg \lgroup1 + \cos \alpha {}^{2} \bigg\rgroup} }$

1 - cos² = sin ²

$\sf \sqrt{ \dfrac{sin {}^{2} \alpha }{ \bigg \lgroup \: 1 + \cos \alpha \bigg \rgroup {}^{2} }}$

$\bf \red{ \sqrt{ \dfrac{ \sin \alpha }{ 1 + \cos \alpha } }}$