MHT CET · Maths · Trigonometric Equations
\(\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}\) has the value
- A \(\frac{\sqrt{5}}{6}\)
- B \(\frac{\sqrt{5}}{6}\)
- C \(\frac{3-\sqrt{5}}{2}\)
- D \(\frac{3+\sqrt{5}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3-\sqrt{5}}{2}\)
Step-by-step Solution
Detailed explanation
Let \(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}=\theta\)
\(\cos 2 \theta=\frac{\sqrt{5}}{3}\)
Now \(\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}\)
\(\begin{aligned}
& =\tan \theta=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}=\sqrt{\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}}=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}} \\
& =\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\frac{3-\sqrt{5}}{2}
\end{aligned}\)
\(\cos 2 \theta=\frac{\sqrt{5}}{3}\)
Now \(\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}\)
\(\begin{aligned}
& =\tan \theta=\sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}=\sqrt{\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}}=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}} \\
& =\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}} \times \frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\frac{3-\sqrt{5}}{2}
\end{aligned}\)
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