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Exercise 3.1 · Q14

Q.Prove that: tan⁡50°=tan⁡40°+2tan⁡10°\tan50°=\tan40°+2\tan10°

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Step 1: tan⁡50°−tan⁡40°=sin⁡(50°−40°)cos⁡50°cos⁡40°=sin⁡10°cos⁡50°cos⁡40°\tan50°-\tan40°=\dfrac{\sin(50°-40°)}{\cos50°\cos40°}=\dfrac{\sin10°}{\cos50°\cos40°}.

Step 2: Now 2cos⁡50°cos⁡40°=cos⁡(50°−40°)+cos⁡(50°+40°)=cos⁡10°+cos⁡90°=cos⁡10°2\cos50°\cos40°=\cos(50°-40°)+\cos(50°+40°)=\cos10°+\cos90°=\cos10°, so cos⁡50°cos⁡40°=cos⁡10°2\cos50°\cos40°=\dfrac{\cos10°}{2}. …

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