Q105 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
Given $\cos \left(\tan ^{-1} x\right)=\sin \left(\cot ^{-1} \dfrac{3}{4}\right)$, then the value of $x$ is
- A$\dfrac{1}{2}$
- B$\dfrac{4}{3}$
- Correct answerC$\dfrac{3}{4}$
- D1
Solution
Let $\theta = \tan^{-1} x$. Then $\tan \theta = x$. Since $\cos \theta = \dfrac{1}{\sqrt{1 + \tan^2 \theta}}$, we have $\cos(\tan^{-1} x) = \dfrac{1}{\sqrt{1 + x^2}}$.
Let $\phi = \cot^{-1} \dfrac{3}{4}$. Then $\cot \phi = \dfrac{3}{4}$, which implies $\tan \phi = \dfrac{4}{3}$.
Using the identity $\sin \phi = \dfrac{1}{\sqrt{1 + \cot^2 \phi}}$, we have $\sin(\cot^{-1} \dfrac{3}{4}) = \dfrac{1}{\sqrt{1 + (3/4)^2}} = \dfrac{1}{\sqrt{1 + 9/16}} = \dfrac{1}{\sqrt{25/16}} = \dfrac{4}{5}$.
Equating the two expressions: $\dfrac{1}{\sqrt{1 + x^2}} = \dfrac{4}{5}$.
Squaring both sides: $\dfrac{1}{1 + x^2} = \dfrac{16}{25}$.
Thus, $1 + x^2 = \dfrac{25}{16}$, which gives $x^2 = \dfrac{25}{16} - 1 = \dfrac{9}{16}$.
Taking the square root, $x = \dfrac{3}{4}$.
Answer: $\dfrac{3}{4}$
Let $\phi = \cot^{-1} \dfrac{3}{4}$. Then $\cot \phi = \dfrac{3}{4}$, which implies $\tan \phi = \dfrac{4}{3}$.
Using the identity $\sin \phi = \dfrac{1}{\sqrt{1 + \cot^2 \phi}}$, we have $\sin(\cot^{-1} \dfrac{3}{4}) = \dfrac{1}{\sqrt{1 + (3/4)^2}} = \dfrac{1}{\sqrt{1 + 9/16}} = \dfrac{1}{\sqrt{25/16}} = \dfrac{4}{5}$.
Equating the two expressions: $\dfrac{1}{\sqrt{1 + x^2}} = \dfrac{4}{5}$.
Squaring both sides: $\dfrac{1}{1 + x^2} = \dfrac{16}{25}$.
Thus, $1 + x^2 = \dfrac{25}{16}$, which gives $x^2 = \dfrac{25}{16} - 1 = \dfrac{9}{16}$.
Taking the square root, $x = \dfrac{3}{4}$.
Answer: $\dfrac{3}{4}$
Hence, the correct answer is option: 'C'.