Q.The value of $\cot\left(\cos^{-1}\frac{7}{25}\right)$ is
(A) $\frac{25}{24}$
(B) $\frac{25}{7}$
(C) $\frac{24}{25}$
(D) $\frac{7}{24}$

Concept: Inverse trigonometric functions and right-triangle interpretation.

Let $\theta = \cos^{-1}\frac{7}{25}$, so $\cos\theta = \frac{7}{25}$ and $\theta \in [0,\pi]$.

Step 1: Since $\cos\theta = \frac{7}{25}$, in a right triangle, adjacent side = 7, hypotenuse = 25.

Step 2: Opposite side = $\sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24$. …

Concept first: When you see an inverse trigonometric function inside another trig function, the cleanest approach is to draw a right triangle. The expression $\cos^{-1}\frac{7}{25}$ means "the angle whose cosine is $\frac{7}{25}$". Call that angle $\theta$. Then $\cos\theta = \frac{7}{25}$. In a right triangle, cosine is adjacent over hypotenuse. So label the adjacent side as 7 and the hypotenuse as 25. The opposite side is then found by Pythagoras. Once you have all three sides, $\cot\theta$ is simply adjacent over opposite.

This avoids messy algebraic manipulation and gives you the answer in seconds.

Step-by-step:

1. Let $\theta = \cos^{-1}\frac{7}{25}$. Then $\cos\theta = \frac{7}{25}$. Since $\cos^{-1}$ returns an angle in $[0, \pi]$, and $\frac{7}{25} > 0$, $\theta$ lies in the first quadrant. So all trig ratios are positive.

2. Draw a right triangle with angle $\theta$. Label the side adjacent to $\theta$ as 7 and the hypotenuse as 25.

3. Use the Pythagorean theorem to find the opposite side:

   $\text{opposite} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24.$ …

Method: Trig ratio of an inverse-trig angle via a right triangle

For $\cot(\cos^{-1}k)$ and similar, treat the inverse function as "the angle whose cosine is $k$," build a right triangle from that ratio, and read off the requested ratio.

Steps

Step 1: Name the angle and record the given ratio.

Let $\theta=\cos^{-1}k$, so $\cos\theta=k$. Since $\cos^{-1}$ returns an angle in $[0,\pi]$, for $k>0$ the angle is acute and all ratios are positive.

Step 2: Label a right triangle.

Write $\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}$ and assign those two sides directly from $k$ (e.g. adjacent $=7$, hypotenuse $=25$).

Step 3: Find the missing side by Pythagoras.

$\text{opposite}=\sqrt{\text{hyp}^2-\text{adj}^2}.$ …

Common Mistakes

Mistake 1: Inverting the ratio and answering $\tfrac{24}{7}$.

Why it's wrong: $\cot\theta=\tfrac{\text{adjacent}}{\text{opposite}}=\tfrac{7}{24}$, not $\tfrac{\text{opposite}}{\text{adjacent}}$. Correct approach: $\cot=\tfrac{1}{\tan}$, so it is adjacent over opposite.

Mistake 2: Confusing $\cot$ with $\csc$ or $\sec$ and using the hypotenuse. …