MHT CET · Maths · Trigonometric Equations
The general solution of \(\tan \theta+\tan 2 \theta=\tan 3 \theta\) is
- A \(\theta=(2 n+1) \frac{\pi}{2}, n \in Z\)
- B \(\theta=n \pi, n \in Z\) or \(\theta=\frac{p \pi}{3}, p \in Z\)
- C \(\theta=\frac{n \pi}{5}, n \in Z\)
- D \(\theta=(2 n-1) \frac{\pi}{3}, n \in Z\)
Answer & Solution
Correct Answer
(B) \(\theta=n \pi, n \in Z\) or \(\theta=\frac{p \pi}{3}, p \in Z\)
Step-by-step Solution
Detailed explanation
\(\tan 3 \theta=\tan (2 \theta+\theta)=\tan \theta+\tan 2 \theta\)
\(\therefore \frac{\tan 2 \theta+\tan \theta}{1-\tan 2 \theta \tan \theta}=\tan \theta+\tan 2 \theta\)
\(\therefore 1-\tan 2 \theta \tan \theta=1 \Rightarrow \tan 2 \theta \tan \theta=0 \Rightarrow \tan \theta=0\)
\(\therefore \theta=\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}\)
\(\therefore \frac{\tan 2 \theta+\tan \theta}{1-\tan 2 \theta \tan \theta}=\tan \theta+\tan 2 \theta\)
\(\therefore 1-\tan 2 \theta \tan \theta=1 \Rightarrow \tan 2 \theta \tan \theta=0 \Rightarrow \tan \theta=0\)
\(\therefore \theta=\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}\)
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