Question

The value of \( x \) such that \( \sin \left( 2 \tan^{-1} \frac{3}{4} \right) = \cos \left( 2 \tan^{-1} x \right) \) is:

• A. \( 7 \)
• B. \( \) (Blank)
• C. \( \frac{1}{7} \)
• D. \( \frac{4}{7} \)

Hint:

For equations involving inverse trigonometric functions, use standard trigonometric identities to simplify the expressions and solve for the unknown.

Answer 1

The Correct Option is C

Step 1: Expressing \( \sin (2\tan^{-1} \theta) \) and \( \cos (2\tan^{-1} \theta) \) Using standard trigonometric identities: \[ \sin (2\tan^{-1} \theta) = \frac{2\theta}{1+\theta^2}, \quad \cos (2\tan^{-1} \theta) = \frac{1-\theta^2}{1+\theta^2}. \]
Step 2: Computing \( \sin(2\tan^{-1} \frac{3}{4}) \) Substituting \( \theta = \frac{3}{4} \): \[ \sin \left( 2 \tan^{-1} \frac{3}{4} \right) = \frac{2 \times \frac{3}{4}}{1 + \left( \frac{3}{4} \right)^2} \] \[ = \frac{\frac{6}{4}}{1 + \frac{9}{16}} \] \[ = \frac{\frac{6}{4}}{\frac{25}{16}} = \frac{6}{4} \times \frac{16}{25} = \frac{96}{100} = \frac{24}{25}. \]
Step 3: Equating with \( \cos(2\tan^{-1} x) \) and Solving for \( x \) Since, \[ \cos(2\tan^{-1} x) = \frac{1-x^2}{1+x^2}, \] we equate: \[ \frac{1 - x^2}{1 + x^2} = \frac{24}{25}. \] Cross multiplying: \[ (1 - x^2) \times 25 = (1 + x^2) \times 24. \] \[ 25 - 25x^2 = 24 + 24x^2. \] \[ 25 - 24 = 25x^2 + 24x^2. \] \[ 1 = 49x^2. \] \[ x^2 = \frac{1}{49}. \] \[ x = \frac{1}{7}. \]
Step 4: Verifying the Correct Option Comparing with the given options, we find: \[ \boxed{\frac{1}{7}}. \] Thus, the correct answer is Option (3).

Answer 2

Given:
\(\sin \left( 2 \tan^{-1} \frac{3}{4} \right) = \cos \left( 2 \tan^{-1} x \right)\)
We know that \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\) and \(\cos(2\theta) = 1 - 2\sin^2(\theta)\).
First, let's find \(\sin \left( 2 \tan^{-1} \frac{3}{4} \right)\):
Let \(\theta = \tan^{-1} \frac{3}{4}\), then \(\tan \theta = \frac{3}{4}\), which implies that the opposite side is 3 and the adjacent side is 4 in a right triangle. Using the Pythagorean theorem, the hypotenuse is \(\sqrt{3^2 + 4^2} = 5\).
Thus, \(\sin \theta = \frac{3}{5}\) and \(\cos \theta = \frac{4}{5}\).
\(\sin \left( 2\tan^{-1} \frac{3}{4} \right) = 2\sin \theta\cos \theta = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}\).
Now, set the equation:
\(\frac{24}{25} = \cos \left( 2\tan^{-1} x \right)\)
Use the formula \(\cos(2\theta) = 1 - 2\sin^2(\theta)\) where \(\theta = \tan^{-1} x\).
\(1 - 2\sin^2(\tan^{-1} x) = \frac{24}{25}\)
\(2\sin^2(\tan^{-1} x) = 1 - \frac{24}{25} = \frac{1}{25}\)
\(\sin^2(\tan^{-1} x) = \frac{1}{50}\)
Now, \(\sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}}\)
Substitute: \(\left(\frac{x}{\sqrt{1+x^2}}\right)^2 = \frac{1}{50}\)
\(\frac{x^2}{1+x^2} = \frac{1}{50}\)
Cross-multiply:
\(50x^2 = 1+x^2\)
\(49x^2 = 1\)
\(x^2 = \frac{1}{49}\)
\(x = \frac{1}{7}\) (since \(x\) is positive)
Therefore, the value of \(x\) is \(\frac{1}{7}\).