If sin B -1-5sin 2 A - B - then tan A - B -tan A is equal toA- 5-3B- 2-3C- 3-2D- 3-5

The correct option is C sin(2A+B)/sin B=5/1 By componendo and dividendo we shall get tan(A+B)/tanA=3/2 ...

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L'Hospital Rule to Remove Indeterminate Form

If sin B =1/5...

Question

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

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

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

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

\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

\frac{s i n A}{s i n B} = \frac{\sqrt{3}}{2} a n d \frac{c o s A}{c o s B} = \frac{\sqrt{5}}{2}, 0 \leq A, B \leq \frac{π}{2},

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

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

\frac{tan A}{tan B}

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