Question

The value of the expression sin \([cot^{-1}(cos (tan^{-1}1))]\) is:

• A. \(\sqrt{\frac{2}{3}}\)
• B. \(\frac{\sqrt{2}}{3}\)
• C. \(\frac{2}{\sqrt{3}}\)
• D. 1

Answer 1

The Correct Option is A

To solve the expression \( \sin \left[ \cot^{-1} \left( \cos (\tan^{-1} 1) \right) \right] \), we follow these steps:

1. Identify \( \tan^{-1} 1 \): Since \( \tan(\frac{\pi}{4}) = 1 \), we have \( \tan^{-1} 1 = \frac{\pi}{4} \).
2. Evaluate \( \cos(\tan^{-1} 1) \):
   In a right triangle where \( \theta = \tan^{-1} 1 \), opposite side = 1, adjacent side = 1 ⇒ hypotenuse = \( \sqrt{2} \). Therefore, \( \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \).
3. Simplify \( \cot^{-1} \left( \frac{1}{\sqrt{2}} \right) \): This implies \( \cot(\theta) = \frac{1}{\sqrt{2}} \). We seek \( \theta \) such that \( \cot(\theta) = \frac{1}{\sqrt{2}} \). For \( \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\), let us assume \( \sin(\theta) = a\) and \(\cos(\theta) = \sqrt{1-a^2} \), then:
   \( \frac{\sqrt{1-a^2}}{a} = \frac{1}{\sqrt{2}} \Rightarrow \sqrt{1-a^2} = \frac{a}{\sqrt{2}} \)
   Squaring both sides: \( 1 - a^2 = \frac{a^2}{2} \)
   ⇒ \( 2(1 - a^2) = a^2 \)
   ⇒ \( 2 - 2a^2 = a^2 \)
   ⇒ \( 2 = 3a^2 \)
   ⇒ \( a^2 = \frac{2}{3} \)
4. Therefore, \( a = \sin(\theta) = \sqrt{\frac{2}{3}} \) and thus the value of the original expression is \(\sqrt{\frac{2}{3}} \).