Q.Solve the following equation $\cos(\tan^{-1}x)=\sin\left(\cot^{-1}\frac{3}{4}\right)$.

Turn each side into an algebraic expression with a right triangle.

Right side: let $\theta=\cot^{-1}\frac{3}{4}$, so $\cot\theta=\frac{3}{4}$ — take adjacent $=3$, opposite $=4$, hypotenuse $=5$. Then $\sin\theta=\frac{4}{5}$. So the equation becomes $\cos(\tan^{-1}x)=\frac{4}{5}$.

Left side: let $\phi=\tan^{-1}x$, so $\tan\phi=x$ — opposite $=x$, adjacent $=1$, hypotenuse $=\sqrt{x^2+1}$. Then $\cos\phi=\frac{1}{\sqrt{x^2+1}}$.

Solve: …

The idea

Whenever a trig function wraps an inverse trig function, draw a right triangle for the inner angle and read off the outer ratio. This converts the equation into ordinary algebra.

Step 1 — Simplify the right-hand side

Let $\theta=\cot^{-1}\frac{3}{4}$, so $\cot\theta=\frac{3}{4}=\frac{\text{adjacent}}{\text{opposite}}$. Take adjacent $=3$, opposite $=4$; then hypotenuse $=\sqrt{3^2+4^2}=5$. So

The equation becomes

Step 2 — Simplify the left-hand side

Let $\phi=\tan^{-1}x$, so $\tan\phi=x=\frac{x}{1}$. Take opposite $=x$, adjacent $=1$; then hypotenuse $=\sqrt{x^2+1}$, and

$\cos\phi=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{1}{\sqrt{x^2+1}}.$ …

Method: Turning $\cos(\tan^{-1}x)$ / $\sin(\cot^{-1}x)$ into algebra

When a trig function wraps an inverse trig function, draw the right triangle of the inner angle and read the outer ratio — this converts the equation into ordinary algebra.

Steps

Step 1: Convert the known side to a number.

For $\sin(\cot^{-1}c)$: put $\cot\theta=c$ (adjacent $c$, opposite $1$, hypotenuse $\sqrt{1+c^2}$), so $\sin\theta=\tfrac{1}{\sqrt{1+c^2}}$ — a fixed value.

Step 2: Convert the unknown side to an expression in $x$.

For $\cos(\tan^{-1}x)$: opposite $x$, adjacent $1$, hypotenuse $\sqrt{1+x^2}$, giving

$\cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^2}}.$ …

Common Mistakes

Mistake 1: Taking $\sin(\cot^{-1}\tfrac34)=\tfrac34$.

Why it's wrong: $\tfrac34$ is the cotangent, not the sine. Correct approach: build the triangle (adjacent $3$, opposite $4$, hypotenuse $5$), so $\sin=\tfrac45$.

Mistake 2: Writing $\cos(\tan^{-1}x)=x$ or $\tfrac{x}{\sqrt{1+x^2}}$.

Why it's wrong: for $\tan\phi=x$ the cosine is adjacent over hypotenuse. Correct approach: $\cos(\tan^{-1}x)=\tfrac{1}{\sqrt{1+x^2}}$. …