KCET · Maths · Matrices
If \( \sin x+\sin y=\frac{1}{2} \) and \( \cos x+\cos y=1 \), then \( \tan (x+y)= \)
- A \( \frac{8}{3} \)
- B \( -\frac{3}{4} \)
- C \( -\frac{8}{3} \)
- D \( \frac{4}{3} \)
Answer & Solution
Correct Answer
(D) \( \frac{4}{3} \)
Step-by-step Solution
Detailed explanation
Given that, \(\sin x+\sin y=\frac{1}{2} \rightarrow(1)\)
\(\cos x+\cos y=1 \rightarrow(2)\)
From Eqs. (1) and (2), we have
\(2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}=\frac{1}{2} \rightarrow(3)\) \(2 \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}=1 \rightarrow(4)\)
Dividing Eq. (3) by Eq. (4), we have
\(\tan \frac{x+y}{2}=\frac{1}{2}\)
We know that \(\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}\)
So, \(\tan (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1-\tan ^{2}\left(\frac{x+y}{2}\right)}\)
\(\Rightarrow \tan (x+y)=\frac{2\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^{2}}=\frac{4}{3}\)
\(\cos x+\cos y=1 \rightarrow(2)\)
From Eqs. (1) and (2), we have
\(2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2}=\frac{1}{2} \rightarrow(3)\) \(2 \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}=1 \rightarrow(4)\)
Dividing Eq. (3) by Eq. (4), we have
\(\tan \frac{x+y}{2}=\frac{1}{2}\)
We know that \(\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}\)
So, \(\tan (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1-\tan ^{2}\left(\frac{x+y}{2}\right)}\)
\(\Rightarrow \tan (x+y)=\frac{2\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^{2}}=\frac{4}{3}\)
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