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Mathematics Question on Trigonometry

For α,β∈(0,π2)\alpha, \beta \in \left( 0, \frac{\pi}{2} \right), let 3sin⁡(α+β)=2sin⁡(α−β)3 \sin(\alpha + \beta) = 2 \sin(\alpha - \beta) and a real number kk be such that tan⁡α=ktan⁡β\tan \alpha = k \tan \beta. Then the value of kk is equal to:

A

−23-\frac{2}{3}

B

–5

C

23\frac{2}{3}

D

5

Answer

–5

Explanation

Solution

Given that

3sin⁡(α+β)=2sin⁡(α−β)3 \sin(\alpha + \beta) = 2 \sin(\alpha - \beta)

and a real number kk such that tan⁡α=ktan⁡β\tan \alpha = k \tan \beta.

Expanding sin⁡(α+β)\sin(\alpha + \beta) and sin⁡(α−β)\sin(\alpha - \beta) using trigonometric identities:

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β,\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta,

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β.\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta.

Substituting these into the given equation:

3(sin⁡αcos⁡β+cos⁡αsin⁡β)=2(sin⁡αcos⁡β−cos⁡αsin⁡β).3(\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2(\sin \alpha \cos \beta - \cos \alpha \sin \beta).

Expanding both sides:

3sin⁡αcos⁡β+3cos⁡αsin⁡β=2sin⁡αcos⁡β−2cos⁡αsin⁡β.3 \sin \alpha \cos \beta + 3 \cos \alpha \sin \beta = 2 \sin \alpha \cos \beta - 2 \cos \alpha \sin \beta.

Collecting terms involving sin⁡αcos⁡β\sin \alpha \cos \beta and cos⁡αsin⁡β\cos \alpha \sin \beta, we get:

3sin⁡αcos⁡β−2sin⁡αcos⁡β=−2cos⁡αsin⁡β−3cos⁡αsin⁡β.3 \sin \alpha \cos \beta - 2 \sin \alpha \cos \beta = -2 \cos \alpha \sin \beta - 3 \cos \alpha \sin \beta.

Simplifying this:

sin⁡αcos⁡β=−5cos⁡αsin⁡β.\sin \alpha \cos \beta = -5 \cos \alpha \sin \beta.

Dividing both sides by cos⁡αcos⁡β\cos \alpha \cos \beta (assuming cos⁡αcos⁡β≠0\cos \alpha \cos \beta \neq 0), we get:

tan⁡α=−5tan⁡β.\tan \alpha = -5 \tan \beta.

Therefore, the value of kk is:

k=−5.k = -5.