Q60 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics

If $\theta=\frac{\pi}{9}$, then $1+27 \tan ^2 \theta-33 \tan ^4 \theta+\tan ^6 \theta=$

  • A
    $3$
  • Correct answer
    B
    $4$
  • C
    $-3$
  • D
    $-11$

Solution

We know, at $\theta=\frac{\pi}{9}$ $\begin{aligned} & \tan 3 \theta=\tan \frac{\pi}{3} \\ & \Rightarrow \frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}=\sqrt{3} \\ & \Rightarrow\left(3 \tan \theta-\tan ^3 \theta\right)^2=3\left(1-3 \tan ^2 \theta\right)^2 \\ & \Rightarrow 9 \tan ^2 \theta+\tan ^6 \theta-6 \tan ^4 \theta \\ & =3\left(1+9 \tan ^4 \theta-6 \tan ^2 \theta\right) \\ & \Rightarrow 9 \tan ^2 \theta+\tan ^6 \theta-6 \tan ^4 \theta \\ & =3+27 \tan ^4 \theta-18 \tan ^2 \theta \\ & \Rightarrow \tan ^6 \theta+27 \tan ^2 \theta-33 \tan ^4 \theta+1=4 \end{aligned}$

Hence, the correct answer is option: 'B'.

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