Q.The value of expression $\tan\left(\frac{\sin^{-1}x+\cos^{-1}x}{2}\right)$, when $x=\frac{\sqrt3}{2}$ is __________.

Concept: Principal Value Domain — $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $x \in [-1,1]$.

Step 1: Use the identity

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

Step 2: Substitute into the expression:

$\tan\left(\frac{\sin^{-1}x+\cos^{-1}x}{2}\right) = \tan\left(\frac{\pi/2}{2}\right) = \tan\left(\frac{\pi}{4}\right)$. …

Why This Works: The Principal Value Domain

The inverse trigonometric functions $\sin^{-1}x$ and $\cos^{-1}x$ are defined on the principal value domains. For $\sin^{-1}x$, the range is $[-\pi/2, \pi/2]$; for $\cos^{-1}x$, the range is $[0, \pi]$.

A beautiful identity connects them: for any $x$ in $[-1, 1]$,

Why? Think geometrically: if an angle has sine $x$, its complement has cosine $x$. The sum of an angle and its complement is always $\pi/2$ radians (90°). This holds regardless of which specific $x$ you pick — it's a constant.

So the expression inside the tangent becomes independent of $x$ — a neat simplification that saves us from plugging in messy values.

Step-by-Step Solution

1. Recall the fundamental identity For any $x \in [-1, 1]$,

This is a standard result from inverse trigonometry, derived from the fact that $\sin\theta = \cos(\pi/2 - \theta)$.

2. Substitute into the given expression The expression is

Using the identity, the numerator becomes $\pi/2$, so:

3. Evaluate the tangent $\tan\left(\frac{\pi}{4}\right) = 1.$ …

Method: Collapse a compound inverse-trig argument with the complementary identity

Steps

Step 1: Spot the complementary sum $\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ inside the argument.

Step 2: Substitute the constant so the argument simplifies: $\dfrac{\pi/2}{2}=\dfrac{\pi}{4}$. …

Common Mistakes

Mistake 1: Evaluating $\sin^{-1}\frac{\sqrt3}{2}$ and $\cos^{-1}\frac{\sqrt3}{2}$ separately and mis-adding.

Why it's wrong: it is unnecessary work that risks an arithmetic slip. Correct approach: the sum is $\frac{\pi}{2}$ regardless of $x$.

Mistake 2: Forgetting to divide by 2. …