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If cosec θ = 5\sqrt{5} and θ is less than 90°, find the value of cot θ - cos θ.

Trigonometrical Ratios

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Answer

Let ABC be a right angle triangle with ∠C = θ and ∠B = 90°.

If cosec θ = √5 and θ is less than 90°, find the value of cot θ - cos θ. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

cosec θ = HypotenusePerpendicular=ACAB\dfrac{\text{Hypotenuse}}{\text{Perpendicular}} = \dfrac{AC}{AB}.

Substituting values we get :

51=ACAB\Rightarrow \dfrac{\sqrt{5}}{1} = \dfrac{AC}{AB}

Let AC = 5\sqrt{5}k and AB = k.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ (5\sqrt{5}k)2 = BC2 + k2

⇒ 5k2 = BC2 + k2

⇒ BC2 = 5k2 - k2

⇒ BC2 = 4k2

⇒ BC = 4k2\sqrt{4\text{k}^2}

⇒ BC = 2k.

cot θ = BasePerpendicular\dfrac{\text{Base}}{\text{Perpendicular}}

= BCAB\dfrac{BC}{AB} = 2kk\dfrac{2k}{k} = 21\dfrac{2}{1}

cos θ = BaseHypotenuse\dfrac{\text{Base}}{\text{Hypotenuse}}

= BCAC\dfrac{BC}{AC} = 2k5k\dfrac{2k}{\sqrt{5}k} = 25\dfrac{2}{\sqrt{5}}

cot θ - cos θ=2125=2525=2(51)5.\therefore \text{cot θ - cos θ} = \dfrac{2}{1} - \dfrac{2}{\sqrt{5}} \\[1em] = \dfrac{2\sqrt{5} - 2}{\sqrt{5}} \\[1em] = \dfrac{2(\sqrt{5} - 1)}{\sqrt{5}}.

Hence, cot θ - cos θ = 2(51)5.\dfrac{2(\sqrt{5} - 1)}{\sqrt{5}}.

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