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Question

Mathematics Question on Trigonometry

If Sin A : Cos A = 4 : 7 , then the value of (7sinA3cosA)(7sinA+2cosA)\frac{(7sinA−3cosA)}{(7sinA+2cosA)} is

A

314\frac{3}{14}

B

32\frac{3}{2}

C

13\frac{1}{3}

D

16\frac{1}{6}

Answer

16\frac{1}{6}

Explanation

Solution

Assume sinA=4and cosA=7sin A = 4 and\ cos A = 7

The expression given in the question, = 7 sinA3 cosA7 sinA+2 cosA\frac{7\ sin A - 3\ cos A}{7\ sin A + 2\ cos A}

From the equation we get = 7(4)3(7)7(4)+2(7)\frac{7(4) - 3(7)}{7(4) + 2(7)}

= 742\frac{7}{42} = 16\frac{1}{6}

The correct option is (D): 16\frac{1}{6}