Question

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Answer 1

Question:-

If Cosecθ = √5, find the value of:-

• (i) 2 - sin²θ - cos²θ
• (ii) 2 + $\sf \dfrac{1}{sin^{2}\theta} - \dfrac{cos^{2}\theta}{sin^{2}\theta}$

Solution:-

We have cosecθ = √5

From here we can find the value of sinθ

We know,

$\sf{cosec\theta = \dfrac{1}{sin\theta}}$

Hence,

We can write

$\sf{\dfrac{1}{sin\theta} = \sqrt{5}}$

$\sf{:\implies sin\theta = \dfrac{1}{\sqrt{5}}}$

Now,

Let us find cosθ in terms of sinθ.

We know,

$\sf{cos\theta = \sqrt{1 - sin^2\theta}}$

Substituting the value of sinθ

= $\sf{cos\theta = \sqrt{1 - \bigg(\dfrac{1}{\sqrt{5}}\bigg)^2}}$

$\sf{:\implies cos\theta = \sqrt{1 - \dfrac{1}{5}}}$

$\sf{:\implies cos\theta = \sqrt{\dfrac{5 - 1}{5}}}$

$\sf{:\implies cos\theta = \sqrt{\dfrac{4}{5}}}$

$\sf{:\implies cos\theta = \dfrac{2}{\sqrt{5}}}$

By rationalising the denominator we get,

$\sf{:\implies cos\theta = \dfrac{2\sqrt{5}}{\sqrt{5}\times\sqrt{5}}}$

$\sf{:\implies cos\theta = \dfrac{2\sqrt{5}}{5}}$

Now,

(i) 2 - sin²θ - cos²θ

⟶ Let us put the value of cosθ and sinθ.

= $\sf{2 - \bigg(\dfrac{1}{\sqrt{5}}\bigg)^2 - \bigg(\dfrac{2\sqrt{5}}{5}\bigg)^2}$

$\sf{:\implies 2 - \bigg[\dfrac{1}{5} + \dfrac{20}{25}\bigg]}$

$\sf{:\implies 2 - \bigg[\dfrac{5 + 20}{25}\bigg]}$

$\sf{:\implies 2 - \bigg[\dfrac{25}{25}\bigg]}$

$\sf{:\implies 2 - 1}$

$\sf{:\implies 1}$

$\sf{\underline{\therefore The\: value\:of\:2 - sin^2\theta - cos^2\theta = 1}}$

(ii) 2 + $\sf{\dfrac{1}{sin^2\theta} - \dfrac{cos^2\theta}{sin^2\theta}}$

= $\sf{2 + \bigg[\dfrac{1}{\bigg(\dfrac{1}{\sqrt{5}}\bigg)^2} - \dfrac{\bigg(\dfrac{2\sqrt{5}}{5}\bigg)^2}{\bigg(\dfrac{1}{\sqrt{5}}\bigg)^2}\bigg]}$

$\sf{:\implies 2 + \bigg[\dfrac{1}{\dfrac{1}{5}} - \dfrac{\dfrac{20}{25}}{\dfrac{1}{5}}\bigg]}$

$\sf{:\implies 2 + \bigg[5 - \dfrac{20}{25}\times 5\bigg]}$

$\sf{:\implies 2 + 5 - 4}$

$\sf{:\implies 7 - 4}$

$\sf{:\implies 3}$

$\sf{\underline{\therefore The\:value\:of\: 2 + \dfrac{1}{sin^2\theta} - \dfrac{cos^2\theta}{sin^2\theta}\: is\: 3}}$

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