Question

The value of x given by \(cos (tan ^{-1}x) = sin (cot^{-1} \ \frac{3}{4})\)is:

• A. \(\frac{1}{4}\)
• B. \(\frac{3}{4}\)
• C. \(\frac{3}{2}\)
• D. 1

Answer 1

The Correct Option is B

To find the value of \(x\) given by \(\cos(\tan^{-1}x) = \sin(\cot^{-1}\frac{3}{4})\), we'll follow these steps:

Step 1: Simplify \(\sin(\cot^{-1}\frac{3}{4})\)

Let \(\theta = \cot^{-1}\frac{3}{4}\). Then \(\cot(\theta) = \frac{3}{4}\).

This implies \(\frac{\text{adjacent}}{\text{opposite}} = \frac{3}{4}\).

Using the Pythagorean identity, the hypotenuse is \(\sqrt{3^2+4^2} = \sqrt{25} = 5\).

Thus, \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}\).

Step 2: Simplify \(\cos(\tan^{-1}x)\)

Let \(\phi = \tan^{-1}x\). Then \(\tan(\phi) = x\), which implies \(\frac{\text{opposite}}{\text{adjacent}} = \frac{x}{1}\).

The hypotenuse is \(\sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}\).

Then, \(\cos(\phi) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2+1}}\).

Step 3: Equate and solve

\[\frac{1}{\sqrt{x^2 + 1}} = \frac{4}{5}\]

Cross-multiply: \(5 = 4\sqrt{x^2 + 1}\).

Divide both sides by 4: \(\frac{5}{4} = \sqrt{x^2 + 1}\).

Square both sides: \(\left(\frac{5}{4}\right)^2 = x^2 + 1\).

\(\frac{25}{16} = x^2 + 1\).

Subtract 1: \(\frac{25}{16} - \frac{16}{16} = x^2\).

\(\frac{9}{16} = x^2\).

Take square root: \(x = \frac{3}{4}\).

Conclusion

The value of \(x\) is \(\frac{3}{4}\), which matches the provided correct answer.