Exams Joint Entrance Examination Mathematics Trigonometric Ratios and Identities Question 196 Question 196 of 298 Medium NUMERICAL
Let S = [ − π , π 2 ] S = \left[-\pi, \dfrac\pi2\right] S = [ − π , 2 π ] (excluding the points where the expression below is undefined), and tan α = 5 \tan\alpha=\sqrt{5} tan α = 5 . Then the number of elements in the set
\[A = \left\{\theta \in S : \tan\theta(1+\sqrt{5}\tan2\theta) = \sqrt{5}-\tan2\theta\right\}\]
is \_\_\_\_\_\_\_\_\_\_\_
Answer: 5
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Rearranging: tan θ + 5 tan θ tan 2 θ + tan 2 θ = 5 ⟹ tan θ + tan 2 θ = 5 ( 1 − tan θ tan 2 θ ) \tan\theta+\sqrt{5}\tan\theta\tan2\theta+\tan2\theta=\sqrt{5} \implies \tan\theta+\tan2\theta = \sqrt{5}(1-\tan\theta\tan2\theta) tan θ + 5 tan θ tan 2 θ + tan 2 θ = 5 ⟹ tan θ + tan 2 θ = 5 ( 1 − tan θ tan 2 θ )
\[\implies \frac{\tan\theta+\tan2\theta}{1-\tan\theta\tan2\theta} = \sqrt{5} = \tan\alpha \implies \tan3\theta = \tan\alpha\]
\[3\theta = n\pi+\alpha \implies \theta = \frac\alpha3+\frac{n\pi}3,\quad n\in\mathbb{Z}\]
Restricting to θ ∈ S = [ − π , π 2 ] \theta \in S = \left[-\pi,\dfrac\pi2\right] θ ∈ S = [ − π , 2 π ] , the acceptable integers are n = 0 , 1 , 2 , 3 , 4 n=0,1,2,3,4 n = 0 , 1 , 2 , 3 , 4 , giving 5 solutions. Answer the question to unlock