If A={n} (A+2 B), then (A+B)=

Correct Answer is : 1+n1−n⋅tan⁡B1+n/1-n · B1−n1+n⋅tanB

Correct Answer is : 1+n1−n⋅tan⁡B1+n/1-n · B1−n1+n⋅tanB

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MHTCET 22 Apr 2025 Shift 2 Paper

Question : 117 of 150

Marks: +1, -0

\sin A=\mathrm{n} \sin (A+2 B)

\tan (A+B)=

Solution:

Concept:

Use componendo-dividendo on the given sine ratio, then apply sine and cosine sum formulas.

Explanation:

Given

\sin A = n \sin(A+2B)

, divide both sides by

\sin A

to get:

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

Applying componendo-dividendo:

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

Now use

\sin X + \sin Y = 2\sin\frac{X+Y}{2}\cos\frac{X-Y}{2}

and

\sin X - \sin Y = 2\cos\frac{X+Y}{2}\sin\frac{X-Y}{2}

with

X = A+2B

and

Y = A

Numerator:

2\sin(A+B)\cos B

Denominator:

2\cos(A+B)\sin B

Therefore:

\frac{2\sin(A+B)\cos B}{2\cos(A+B)\sin B} = \frac{1+n}{1-n}

Simplify to:

\frac{\tan(A+B)}{\tan B} = \frac{1+n}{1-n}

So:

\tan(A+B) = \left(\frac{1+n}{1-n}\right)\tan B

Answer:

\tan(A+B) = \left(\frac{1+n}{1-n}\right)\tan B

, which matches option D.

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