Q.The value of $\cos(\sin^{-1}x+\cos^{-1}x)$, $|x|\le1$ is __________.

The key idea is that $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $|x| \le 1$.

Step 1: Recall the identity:

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$.

Step 2: Substitute into the given expression: …

Why this works

The key insight is that $\sin^{-1}x$ and $\cos^{-1}x$ are complementary angles — their sum is always $\frac{\pi}{2}$ for any $x$ in $[-1, 1]$. This is not a coincidence; it follows directly from the definitions of inverse trigonometric functions.

Think of a right triangle: if an angle has sine $x$, its complement has cosine $x$. The inverse functions simply "undo" the trig functions, so the sum of the two inverse functions gives the sum of two complementary angles.

Once you know the sum is constant, the cosine of that constant is trivial to compute.

Step-by-step solution

1. Recall the fundamental identity For any $x$ with $|x| \le 1$, we have:

This is a standard result — it holds because $\sin\theta = \cos(\frac{\pi}{2} - \theta)$, so if $\sin^{-1}x = \theta$, then $\cos^{-1}x = \frac{\pi}{2} - \theta$.

2. Substitute into the given expression The problem asks for $\cos(\sin^{-1}x + \cos^{-1}x)$. Using the identity:

3. Evaluate the cosine We know $\cos(\pi/2) = 0$. Therefore: $\cos(\sin^{-1}x + \cos^{-1}x) = 0$ …

Method: Use the complementary identity $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$

Steps

Step 1: Recognise a complementary pair.

For any $x\in[-1,1]$, $\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ — a constant, independent of $x$. …

Common Mistakes

Mistake 1: Treating the sum as depending on $x$.

Why it's wrong: $\sin^{-1}x+\cos^{-1}x$ is always $\frac{\pi}{2}$ on $[-1,1]$, so the outer value is fixed. Correct approach: substitute $\frac{\pi}{2}$ directly.

Mistake 2: Expanding with the cosine addition formula and slipping. …