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A and B are positive acute angles satisfying the equations `3cos^(2)A+2cos^(2)B=4and (3sinA)/(sinB)=(2cosB)/(cosA), then A+2B` is equal toA. `pi//4`B. `pi//3`C. `pi//6`D. `pi//2` |
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Answer» Correct Answer - D From the given relation, we have `2 cos^(2) B-1=3(1-cos^(2)A)` or `cos 2B = 3 sin^(2)A` …(1) `(3 sin A)/(sin B)=(2 cos B)/(cos A)` `rArr (3)/(2) sin 2A = sin 2B` `rArr 3 sin 2A = 2 sin 2B` …(2) Now cos (A + 2B) `= cos A cos 2B - sin A sin 2B` `= cos A(3 sin^(2)A)-sin A.(3)/(2)sin 2A` `= 3 cos A sin^(2)A-3 sin^(2)A cos A = 0` `therefore A+2B=90^(@)` |
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