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e If 1+ sin® 0 = 3 sin cos , then prove that tan®=1ortan 0= |
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Answer» 1+sin^2 theta=3 sintheta cos theta (we know that sin^2 theta + cos^2 theta =1) = ( sin^2 theta + cos^2 theta ) + sin ^2 theta = 3 sin theta cos theta = sin^2 theta + cos^2 theta + sin ^2 theta = 3 sin theta cos theta = cos^2 theta + 2 sin^2 theta = 3 sin theta cos theta On dividing by cos^2 theta, we get = 1 + 2 tan^2 theta = 3 tan theta Let tan theta = b 2b^2 - 3b + 1 = 0 = (2b-1)(b-1) = 0 b = 1 or 1/2 So, tan theta = 1 or 1/2. |
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