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Mathematics Question on Trigonometric Functions

If sin⁡x−sin⁡y=12\sin x- \sin y = \frac {1}{2} and cos⁡x−cos⁡y=1\cos x- \cos y = 1, then tan⁡(x+y)\tan(x + y) is equal to

A

38\frac {3}{8}

B

−38-\frac{3}{8}

C

43\frac {4}{3}

D

−43-\frac {4}{3}

Answer

43\frac {4}{3}

Explanation

Solution

Given,
sin⁡x−sin⁡y=12        ...(i)\sin x-\sin y=\frac{1}{2}\,\,\,\,\,\,\,\,...(i)
and cos⁡x−cos⁡y=1        ...(ii) \cos x-\cos y=1 \,\,\,\,\,\,\,\,...(ii)
⇒2cos⁡x+y2⋅sin⁡x−y2=12        ...(iii)\Rightarrow 2 \cos \frac{x+y}{2} \cdot \sin \frac{x-y}{2}=\frac{1}{2} \,\,\,\,\,\,\,\,...(iii)
and −2sin⁡x+y2⋅sin⁡x−y2=1        ...(iv)-2 \sin \frac{x+y}{2} \cdot \sin \frac{x-y}{2}=1\,\,\,\,\,\,\,\,...(iv)
On dividing E (iv) by E (iii), we get
−tan⁡(x+y2)=2-\tan \left(\frac{x+y}{2}\right)=2
⇒tan⁡(x+y2)=−2        ...(v)\Rightarrow \tan \left(\frac{x+y}{2}\right)=-2\,\,\,\,\,\,\,\,...(v)
Now, tan⁡(x+y)=2tan⁡(x+y2)1−tan⁡2(x+y2)\tan (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1-\tan ^{2}\left(\frac{x+y}{2}\right)}
(∵tan⁡2 θ=2tan⁡θ1−tan⁡2θ)\left(\because \tan 2\,\theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)
=2(−2)1−(−2)2      =\frac{2(-2)}{1-(-2)^{2}}\,\,\,\,\,\,[using E (v)]
=−41−4=−4−3=43=\frac{-4}{1-4}=\frac{-4}{-3}=\frac{4}{3}