If cos Ɵ = √3/2 and 0° < Ɵ < 90°, then the value of cosec (Ɵ �

1.	If cos Ɵ = √3/2 and 0° < Ɵ < 90°, then the value of cosec (Ɵ – 15°) is1). 12). √2 + √63). 1/√34). √2
Answer» If COS ? = √3/2 ⇒ ? = 30° ∴ COSEC (? – 15°) = cosec (30° – 15°) = cosec 15° = 1/(sin 15°) Since sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° (? sin (A – B) = sin A cos B – cos A sin B) ∴ sin 15° $(= \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 2 }} \times \frac{1}{2} = \frac{{\sqrt 3 \;-1}}{{2\sqrt 2 }})$ ⇒ cosec 15° = $(\frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} = \frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} = \;\frac{{2\sqrt 6 + 2\sqrt 2 }}{{3 - 1}} = \sqrt 2 + \sqrt 6 )$

If cos Ɵ = √3/2 and 0° < Ɵ < 90°, then the value of cosec (Ɵ – 15°) is1). 12). √2 + √63). 1/√34). √2

Answer»

If COS ? = √3/2

⇒ ? = 30°

∴ COSEC (? – 15°) = cosec (30° – 15°) = cosec 15° = 1/(sin 15°)

Since sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30°

(? sin (A – B) = sin A cos B – cos A sin B)

∴ sin 15° $(= \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 2 }} \times \frac{1}{2} = \frac{{\sqrt 3 \;-1}}{{2\sqrt 2 }})$

⇒ cosec 15° = $(\frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} = \frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} = \;\frac{{2\sqrt 6 + 2\sqrt 2 }}{{3 - 1}} = \sqrt 2 + \sqrt 6 )$

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