Question

solve kar do yaar plzzz

tennetiraj bhaiya ​

Answer 1

${\huge{\underbrace{\rm{Answer\checkmark}}}}$

★ First One -

$\rm \longrightarrow \: \sqrt{ \dfrac{1 + \cos( \theta) }{1 - \cos( \theta) } } + \sqrt{ \dfrac{1 - \cos( \theta) }{1 + \cos( \theta) } }$

Rationalize the denominator ...

$\implies \rm \: \sqrt{ \frac{(1 + \cos( \theta) )(1 - \cos( \theta) )}{(1 - \cos( \theta) )(1 + \cos( \theta)) } } + \sqrt{ \frac{(1 - \cos( \theta))(1 - \cos( \theta)) }{(1 + \cos( \theta))(1 - \cos( \theta)) } }$

$\implies \rm \: \dfrac{1 + \cos( \theta) }{ \sqrt{1 - { \cos }^{2} \theta} } + \dfrac{1 - \cos( \theta) }{ \sqrt{1 - { \cos }^{2} \theta} }$

$\implies \rm \: \dfrac{1 + \cos( \theta) }{ \sqrt{ { \sin}^{2} \theta } } + \dfrac{1 - \cos( \theta) }{ \sqrt{ { \sin }^{2} \theta } }$

$\implies \rm \: \dfrac{1 + \cos( \theta) }{ \sin( \theta) } + \dfrac{1 - \cos( \theta) }{ \sin( \theta) }$

$\implies \rm \: \dfrac{1 + \cos( \theta) + 1 - \cos( \theta) }{ \sin( \theta) }$

$\implies \rm \: \dfrac{2}{ \sin( \theta) }$

$\implies \rm \: 2 \sin( \theta)$

$\therefore\: \rm \: hence \: proved$

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★ Formula Used -

$\mapsto \rm \: (x + y)(x - y) = {x}^{2} - {y}^{2}$

$\mapsto \rm \: 1 - {cos}^{2} \theta \: = \: {sin}^{2} \theta$

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★ Second One -

$\rm \: answer \: in \: attachment$

★ Formula Used -

$\mapsto \rm \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$

$\mapsto \rm \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy)$

$\mapsto \rm \: {cos}^{2} \theta + {sin}^{2} \theta = 1$

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Thanks for points...xD

Answer 2

Answer:

★ First One :-

{1 + \cos( \theta) }{1 - \cos( \theta) } } + {1 - \cos( \theta) }{1 + \cos( \theta) } }⟶1−cos(θ)1+cos(θ)+1+cos(θ)1−cos(θ)

Rationalize the denominator ...

{(1 + \cos( \theta) )(1 - \cos( \theta) )}{(1 - \cos( \theta) )(1 + \cos( \theta)) } } + \sqrt{ \frac{(1 - \cos( \theta))(1 - \cos( \theta)) }{(1 + \cos( \theta))(1 - \cos( \theta)) } }⟹(1−cos(θ))(1+cos(θ))(1+cos(θ))(1−cos(θ))+(1+cos(θ))(1−cos(θ))(1−cos(θ))(1−cos(θ))

{1 + \cos( \theta) }{ \sqrt{1 - { \cos }^{2} \theta} } + \dfrac{1 - \cos( \theta) }{ \sqrt{1 - { \cos }^{2} \theta} }⟹1−cos2θ1+cos(θ)+1−cos2θ1−cos(θ)

{1 + \cos( \theta) }{ \sqrt{ { \sin}^{2} \theta } } + \dfrac{1 - \cos( \theta) }{ \sqrt{ { \sin }^{2} \theta } }⟹sin2θ1+cos(θ)+sin2θ1−cos(θ)

{1 + \cos( \theta) }{ \sin( \theta) } + {1 - \cos( \theta) }{ \sin( \theta) }⟹sin(θ)1+cos(θ)+sin(θ)1−cos(θ)

{1 + \cos( \theta) + 1 - \cos( \theta) }{ \sin( \theta) }⟹sin(θ)1+cos(θ)+1−cos(θ)

{2}{ \sin( \theta) }⟹sin(θ)2

2 \sin( \theta)⟹2sin(θ)

therefore,hence proved∴

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★ Formula Used :-

(x + y)(x - y) = {x}^{2} - {y}^{2}

↦(x+y)(x−y)=x2−y2

1 - {cos}^{2} theta = {sin}^{2} \theta

↦1−cos2θ=sin2θ

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★ Second One :-

Answer is in attachment.

★ Formula Used :-

{x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)

↦x3+y3=(x+y)(x2+y2−xy)

{x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy)

↦x3−y3=(x−y)(x2+y2+xy)

{cos}^{2} \theta + {sin}^{2} \theta = 1

↦cos2θ+sin2θ=1

HOPE IT WILL HELP YOU..