Question

If \(x = \sin(2\tan^{-1}2)\), \(y = \sin\left(\frac{1}{2}\tan^{-1}\frac{4}{3}\right)\), then:

• A. \(x = 1 - y\)
• B. \(x^2 = 1 - y\)
• C. \(x^2 = 1 + y\)
• D. \(y^2 = 1 - x\)

Hint:

Convert \(\tan^{-1}\) values into right triangles to quickly get \(\sin\theta, \cos\theta\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We evaluate \(x\) and \(y\) using trigonometric substitutions and double/half-angle formulas to find a relationship between them.
Step 2: Key Formula or Approach:
1. \(\sin(2\theta) = \frac{2\tan\theta}{1+\tan^{2}\theta}\)
2. \(\sin(\alpha/2) = \sqrt{\frac{1-\cos\alpha}{2}}\)
: Detailed Explanation:
1. For \(x = \sin(2 \tan^{-1} 2)\):
Let \(\theta = \tan^{-1} 2 \implies \tan\theta = 2\).
\[ x = \sin(2\theta) = \frac{2(2)}{1+2^{2}} = \frac{4}{5} \]
2. For \(y = \sin(\frac{1}{2} \tan^{-1} \frac{4}{3})\):
Let \(\alpha = \tan^{-1} \frac{4}{3} \implies \tan\alpha = \frac{4}{3}\).
Since \(\tan\alpha = 4/3\), then \(\cos\alpha = 3/5\) (using a \(3-4-5\) triangle).
\[ y = \sin(\alpha/2) = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} \]
3. Comparing \(x\) and \(y\):
\[ y^{2} = \frac{1}{5} \]
\[ 1 - x = 1 - \frac{4}{5} = \frac{1}{5} \]
Thus, \(y^{2} = 1 - x\).
Step 3: Final Answer:
The relation is \(y^{2} = 1 - x\).