Question

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Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\:ycos\theta = x}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{1}{cos\theta } = \dfrac{y}{x}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{1 + cos\theta }{1 - cos\theta } = \dfrac{y + x}{y - x}$

On rationalizing the LHS, we get

$\rm :\longmapsto\:\dfrac{1 + cos\theta }{1 - cos\theta } \times \dfrac{1 + cos\theta }{1 - cos\theta } = \dfrac{y + x}{y - x}$

$\rm :\longmapsto\:\dfrac{ {(1 + cos\theta )}^{2} }{1 - {cos}^{2} \theta } = \dfrac{y + x}{y - x}$

$\red{\bigg \{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

and

We know,

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

Using this, we get

$\rm :\longmapsto\:\dfrac{ {(1 + cos\theta )}^{2} }{{sin}^{2} \theta } = \dfrac{y + x}{y - x}$

$\rm :\longmapsto\: {\bigg(\dfrac{1 + cos\theta }{sin\theta } \bigg) }^{2} = \dfrac{y + x}{y - x}$

$\rm :\longmapsto\: {\bigg(\dfrac{1}{sin\theta } + \dfrac{cos\theta }{sin\theta } \bigg) }^{2} = \dfrac{y + x}{y - x}$

$\rm :\longmapsto\: {(cosec\theta + cot\theta )}^{2} = \dfrac{y + x}{y - x}$

$\bf\implies \:cosec\theta + cot\theta = \sqrt{\dfrac{y + x}{y - x} }$

Hence, Proved

Results used :-

1. Componendo and Dividendo

$\boxed{ \rm{ \dfrac{a}{b} = \dfrac{c}{d} \rm \: \implies\:\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}}}$

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

$\boxed{ \rm{ \frac{cos\theta }{sin\theta } = cot\theta }}$

$\boxed{ \rm{ \frac{1}{sin\theta } = cosec\theta }}$

Additional Information :-

$\boxed{ \rm{ {sec}^{2} \theta - {tan}^{2} \theta = 1}}$

$\boxed{ \rm{ {cosec}^{2} \theta - {cot}^{2} \theta = 1}}$

$\boxed{ \rm{ tan\theta = \frac{sin\theta }{cos\theta } = \frac{1}{cot\theta }}}$