Given that α+β=90^{circ}, show that √{cos α operatorname{cosec} β-cos α sin β}=sin α

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8. Introduction to Trigonometry

easy

Given that $\alpha+\beta=90^{\circ},$ show that

$\sqrt{\cos \alpha \operatorname{cosec} \beta-\cos \alpha \sin \beta}=\sin \alpha$

Option A

Option B

Option C

Option D

Solution

$\sqrt{\cos \alpha \operatorname{cosec} \beta-\cos \alpha \sin \beta}=\sqrt{\cos \alpha \operatorname{cosec}\left(90^{\circ}-\alpha\right)-\cos \alpha \sin \left(90^{\circ}-\alpha\right)}$ $\left[\right.$ Given $\left.\alpha+\beta=90^{\circ}\right]$

$=\sqrt{\cos \alpha \sec \alpha-\cos \alpha \cos \alpha}$

$=\sqrt{1 \cos ^{2}}$

$=\sin \alpha$

Standard 10

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Given that $\alpha+\beta=90^{\circ},$ show that $\sqrt{\cos \alpha \operatorname{cosec} \beta-\cos \alpha \sin \beta}=\sin \alpha$

Solution

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Given that $\alpha+\beta=90^{\circ},$ show that

$\sqrt{\cos \alpha \operatorname{cosec} \beta-\cos \alpha \sin \beta}=\sin \alpha$