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3 = 5sinf —3cosOनए 550 0+ 3९०५9 |
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Answer» We are given 3 cot x = 4, or cot x = 4/3. Now consider a right angled triangle, ABC, with BC as the base , right angle at and A the upper vertex. Thus <ABC = x. So BC=4, AC=3 and AB=5. Then sin x = 3/5 and cos x = 4/5 (1) (5 sin x + 3 cos x)/(5 sin x - 3 cos x) … (2). Put the values of sin x and cos x from (1) it (2) to get [(5*3/5) +(3*4/5)]/[(5*3/5) -(3*4/5)], or = (15+12)/(15–12) [As 5 is common both in the numerator and denominator it can be ignored] = 27/3 = 9 Like my answer if you find it useful! |
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