Question

If \(\tan(\sec^{-1}x) = \sin\left(\cos^{-1}\frac{1}{\sqrt{5}}\right)\), then \(x\) is equal to

• A. \(\pm \frac{3}{\sqrt{5}}\)
• B. \(\pm \frac{\sqrt{5}}{3}\)
• C. \(\pm \sqrt{\frac{3}{5}}\)
• D. None of these

Hint:

\(\tan(\sec^{-1}x) = \sqrt{x^2-1}\) for \(x \ge 1\).

Answer 1

The Correct Option is A

To solve the given problem, we need to evaluate both sides of the equation and find the value of \(x\).

We begin by evaluating the left-hand side of the equation \( \tan(\sec^{-1}x) \):

1. \(\sec^{-1}x\) represents the angle whose secant is \(x\). Let this angle be \(\theta\), then: \(\sec \theta = x\)
2. We know that: \(\sec \theta = \frac{1}{\cos \theta}\), hence \( \cos \theta = \frac{1}{x} \)
3. Then, use the identity for tangent: \(\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{x^2 - 1}\)
4. Thus, \(\tan(\sec^{-1}x) = \sqrt{x^2 - 1}\)

Next, evaluate the right-hand side of the equation \(\sin\left(\cos^{-1}\frac{1}{\sqrt{5}}\right)\):

1. \(\cos^{-1}\frac{1}{\sqrt{5}}\) represents the angle whose cosine is \(\frac{1}{\sqrt{5}}\). Let this angle be \(\phi\), then: \(\cos \phi = \frac{1}{\sqrt{5}}\)
2. We use the identity for sine: \(\sin \phi = \sqrt{1 - \cos^2 \phi}\)
3. Plug \(\cos \phi = \frac{1}{\sqrt{5}}\) into the identity: \(\sin \phi = \sqrt{1 - \left(\frac{1}{\sqrt{5}}\right)^2} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}\)
4. Thus, \(\sin\left(\cos^{-1}\frac{1}{\sqrt{5}}\right) = \frac{2}{\sqrt{5}}\)

Now, set the two expressions equal:

1. \(\sqrt{x^2 - 1} = \frac{2}{\sqrt{5}}\)
2. Square both sides to eliminate the square root: \(x^2 - 1 = \left(\frac{2}{\sqrt{5}}\right)^2 = \frac{4}{5}\)
3. Add 1 to both sides: \(x^2 = \frac{4}{5} + 1 = \frac{4}{5} + \frac{5}{5} = \frac{9}{5}\)
4. Solve for \(x\): \(x = \pm \sqrt{\frac{9}{5}} = \pm \frac{3}{\sqrt{5}}\)

Hence, the value of \(x\) is \( \pm \frac{3}{\sqrt{5}} \), which matches with the provided correct option.