If sin B- 1-5sin2A - B- then tanA-B-tanA is equal to

Sin(2A+B)/sin B=5/1 By componendo and dividendo we shall get tan(A+B)/tanA=3/2 ...

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Sum of Trigonometric Ratios in Terms of Their Product

If sin B= 1/5...

Question

If $sin B = \frac{1}{5} sin (2 A + B),$ then $\frac{t a n (A + B)}{t a n A}$ is equal to

\frac{5}{3}

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\frac{2}{3}

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\frac{3}{2}

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\frac{3}{5}

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\frac{1}{5} s i n (2 A + B), t h e n \frac{t a n (A + B)}{t a n A}

sin B = \frac{1}{5} sin (2 A + B)

tan (A + B) = k tan A

k

sin B = \frac{1}{5} sin (2 A + B),

\frac{t a n (A + B)}{t a n A}

\frac{sin A}{sin B} = \frac{\sqrt{3}}{2}

\frac{cos A}{cos B} = \frac{\sqrt{5}}{2}, 0 < A, B < π / 2,

tan A + tan B

sin B = 3 sin (2 A + B)

2 tan A + tan (A + B) =

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