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Miscellaneous Exercise 3 (II) · Q118

Q.Prove the following: 3 cosec 20°−sec⁡20°=4\sqrt3\,\text{cosec}\,20°-\sec20°=4

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Step 1: 3 cosec 20°−sec⁡20°=3sin⁡20°−1cos⁡20°=3cos⁡20°−sin⁡20°sin⁡20°cos⁡20°\sqrt3\,\text{cosec}\,20°-\sec20°=\dfrac{\sqrt3}{\sin20°}-\dfrac{1}{\cos20°}=\dfrac{\sqrt3\cos20°-\sin20°}{\sin20°\cos20°}.

Step 2: Numerator: 3cos⁡20°−sin⁡20°=2(32cos⁡20°−12sin⁡20°)=2(sin⁡60°cos⁡20°−cos⁡60°sin⁡20°)=2sin⁡(60°−20°)=2sin⁡40°\sqrt3\cos20°-\sin20°=2\left(\dfrac{\sqrt3}{2}\cos20°-\dfrac12\sin20°\right)=2(\sin60°\cos20°-\cos60°\sin20°)=2\sin(60°-20°)=2\sin40°. …

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