Trigonometry
Trigonometry
Allen Star Batch
Grade 11
Question:
The equation $(1-\tan\theta)(1+\tan\theta)\sec^2\theta+2\tan^2\theta=0$ has:
No solution in the interval $\left(-\frac{\pi}{2},0\right)$
Two solutions in the interval $\left(-\frac{\pi}{2},0\right)$
No solution in the interval $\left(0,\frac{\pi}{2}\right)$
Two solutions in the interval $\left(0,\frac{\pi}{2}\right)$
Step-by-Step Solution
Key Concept: Convert transcendental equations to algebraic form and solve using graphical intersection methods.
The equation $(1-\tan^2\theta)(1+\tan^2\theta)\sec^2\theta + 2^{\sin^2\theta} = 0$ simplifies to $(1-\tan^2\theta) + 2^{\sin^2\theta} = 0$. Substituting $t = \tan^2\theta$ gives $2^t = t^2 - 1$. Solving graphically, $t = 3$ is a solution, yielding $\tan^2\theta = 3$ or $\tan\theta = \pm\sqrt{3}$, so $\theta = \pm\frac{\pi}{3}$. Since another root exists after $t=3$, two solutions lie in $[0, \frac{\pi}{2})$ and $[-\frac{\pi}{2}, 0)$.
Correct Answer: 2,4