Solve the equation cos(tan⁻¹ x) = sin(cot⁻¹ 3/4) | NCERT EXEMPLAR Class 12 Maths Ex 2.3 Q11

Solution

We have:

\[ \cos(\tan^{-1}x) = \sin(\cot^{-1}\frac{3}{4}) \]

Using inverse trigonometric identities:

\[ \cos(\tan^{-1}x) = \frac{1}{\sqrt{x^2 + 1}}, \quad \sin(\cot^{-1}\frac{3}{4}) = \frac{4}{5} \]

Let \(\tan^{-1}x = \theta_1 \Rightarrow \tan \theta_1 = x\), then

\[ \cos \theta_1 = \frac{1}{\sqrt{x^2 + 1}} \Rightarrow \theta_1 = \cos^{-1} \frac{1}{\sqrt{x^2 + 1}} \]

And \(\cot^{-1}\frac{3}{4} = \theta_2 \Rightarrow \cot \theta_2 = \frac{3}{4}\), then

\[ \sin \theta_2 = \frac{4}{5} \Rightarrow \theta_2 = \sin^{-1} \frac{4}{5} \]

Equating both sides:

\[ \frac{1}{\sqrt{x^2 + 1}} = \frac{4}{5}, \quad \text{(since } \sin(\sin^{-1} y) = y, \; y \in [-1,1]) \]

Squaring both sides:

\[ 16(x^2 + 1) = 25 \Rightarrow 16x^2 = 9 \Rightarrow x^2 = \left(\frac{3}{4}\right)^2 \]

Therefore:

\[ \boxed{x = \pm \frac{3}{4}} \]