Question

$In(0,\pi ),\text{the number of solutions of the equation}tan\theta +tan2\theta +tan3\theta =tan\theta tan2\theta tan3\theta is$

• A. 7
• B. 5
• C. 4
• D. 2

Answer 1

Answer: (D)

2

$Givenequation:tan\theta +tan2\theta +tan3\theta =tan\theta tan2\theta tan3\theta \Rightarrow tan\theta +tan2\theta =-tan3\theta +tan\theta tan2\theta tan3\theta \Rightarrow tan\theta +tan2\theta =-tan3\theta (1-tan\theta tan2\theta )\Rightarrow \frac{tan\theta +tan2\theta }{1-tan\theta tan2\theta }=-tan3\theta \Rightarrow tan(\theta +2\theta )=-tan3\theta \Rightarrow tan3\theta =-tan3\theta \Rightarrow 2tan3\theta =0\Rightarrow tan3\theta =0\Rightarrow 3\theta =n\pi \Rightarrow \theta =\frac{n\pi }{3}Now,\theta =\frac{\pi }{3},n=1\theta =\frac{2\pi }{3},n=2\theta =\frac{3\pi }{3}={180}^{\circ },\text{which is not possible as it is not in the interval}(0,\pi )\text{Hence,the number of solutions of the given equation is 2.}$