Q.If $\cos^{-1}x>\sin^{-1}x$, then
(A) $\frac{1}{\sqrt2}<x\le1$
(B) $0\le x<\frac{1}{\sqrt2}$
(C) $-1\le x<\frac{1}{\sqrt2}$
(D) $x>0$

Concept: Range and monotonicity of inverse trigonometric functions.

$\cos^{-1}x$ is decreasing on $[-1,1]$, $\sin^{-1}x$ is increasing on $[-1,1]$, and they satisfy $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$.

Step 1: Rewrite the inequality using the identity.

$\cos^{-1}x > \sin^{-1}x$ implies $\frac{\pi}{2} - \sin^{-1}x > \sin^{-1}x$, so $\frac{\pi}{2} > 2\sin^{-1}x$, i.e., $\sin^{-1}x < \frac{\pi}{4}$. …

The key here is to avoid plugging in random values. Instead, use the fundamental relationship between inverse sine and cosine.

For any $x$ in $[-1, 1]$, we have the identity:

This is true because $\sin\theta$ and $\cos\theta$ are complementary angles — if $\sin\theta = x$, then $\cos\left(\frac{\pi}{2} - \theta\right) = x$, so the inverse functions are complementary.

Now, the inequality given is:

Replace $\sin^{-1}x$ using the identity:

Add $\cos^{-1}x$ to both sides:

Divide by 2:

So the problem reduces to: for which $x$ in the domain $[-1, 1]$ is $\cos^{-1}x$ greater than $\frac{\pi}{4}$?

Now recall the graph of $y = \cos^{-1}x$: it is a decreasing function from $x = -1$ (where $\cos^{-1}(-1) = \pi$) to $x = 1$ (where $\cos^{-1}(1) = 0$). So $\cos^{-1}x > \frac{\pi}{4}$ means $x$ must be less than the value where $\cos^{-1}x = \frac{\pi}{4}$.

What is that value? Solve:

Since $\cos^{-1}x$ is decreasing, for $x < \frac{1}{\sqrt{2}}$, we have $\cos^{-1}x > \frac{\pi}{4}$. For $x = \frac{1}{\sqrt{2}}$, equality holds, so it's excluded. For $x > \frac{1}{\sqrt{2}}$, the inequality reverses. …

Method: Solving inequalities between $\cos^{-1}x$ and $\sin^{-1}x$

Inequalities comparing two inverse functions of the same variable are best solved by using the complementary identity to reduce to one function, then applying its monotonicity — and never forgetting the shared domain.

Steps

Step 1: Reduce to a single inverse function.

Use $\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ to replace one function. From $\cos^{-1}x>\sin^{-1}x$:

Step 2: Find the boundary value.

Solve the equality $\cos^{-1}x=\tfrac{\pi}{4}$, giving $x=\cos\tfrac{\pi}{4}=\tfrac{1}{\sqrt2}$.

Step 3: Apply monotonicity to fix the direction. …

Common Mistakes

Mistake 1: Getting the inequality direction backwards.

Why it's wrong: $\cos^{-1}x$ is a decreasing function, so $\cos^{-1}x>\tfrac{\pi}{4}$ means $x<\cos\tfrac{\pi}{4}=\tfrac{1}{\sqrt2}$, not $x>\tfrac{1}{\sqrt2}$. Correct approach: account for monotonicity when converting an angle inequality to an $x$ inequality.

Mistake 2: Dropping the lower bound and answering $0\le x<\tfrac{1}{\sqrt2}$. …