The value of cot ≤ft[ cos^-1 ≤ft( 7/25 ) ] is

The correct answer is Option (4) → $\frac{7}{24}$ ##

We have, $\cot \left[ \cos^{-1} \left( \frac{7}{25} \right) \right]$

Let $\cos^{-1} \frac{7}{25} = x$

$⇒\cos x = \frac{7}{25}$

$∴\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - \left( \frac{7}{25} \right)^2}$

$= \sqrt{\frac{625 - 49}{625}} = \frac{24}{25}$

and $\cot x = \frac{\cos x}{\sin x} = \frac{\frac{7}{25}}{\frac{24}{25}} = \frac{7}{24} \dots(i)$

$⇒x = \cot^{-1} \left( \frac{7}{24} \right) = \cos^{-1} \left( \frac{7}{25} \right)$

$∴\cot \left( \cos^{-1} \frac{7}{25} \right) = \cot \left( \cot^{-1} \frac{7}{24} \right) = \frac{7}{24} \quad \left[ ∵\cot^{-1} \frac{7}{24} = \cos^{-1} \frac{7}{25} \right]$