Trigonometry & Inverse Trigonometry
Trig Ratios Functions Identities
nta_abhyas_2025
Grade None
Question:
If $\pi < \theta < \frac{3\pi}{2}$ and $\cos \theta = -\frac{3}{5}$, then $\tan \left(\frac{\theta}{2}\right)$ is equal to
\frac{-5+1}{4}
\frac{\sqrt{5}+1}{4}
\frac{-\sqrt{5}-1}{4}
\frac{\sqrt{5}-1}{4}
Step-by-Step Solution
Key Concept: Half-angle formulas $\cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}$ and $\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}$ are applied to find $\tan\frac{\theta}{2}$.
Given $\cos\theta = -\frac{4}{5}$, we use the half-angle formula $\cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2} = \frac{1 - \frac{4}{5}}{2} = \frac{\frac{1}{5}}{2} = \frac{1}{10}$. Since $\cos\frac{\theta}{2} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$, and using $\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2} = \frac{9}{10}$, we get $\sin\frac{\theta}{2} = \frac{3}{\sqrt{10}}$. Therefore $\tan\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = 3$, and $\tan^2\frac{\theta}{2} = 9 = \left(\frac{\sqrt{5}+1}{\sqrt{5}-1}\right)^2$ simplifies to the answer.
Correct Answer: 2