Q.If $\cos\left(\sin^{-1}\frac{2}{5}+\cos^{-1}x\right)=0$, then $x$ is equal to
(A) $\frac{1}{5}$
(B) $\frac{2}{5}$
(C) $0$
(D) $1$

Concept: Inverse trigonometric functions and the condition $\cos\theta = 0 \implies \theta = \frac{\pi}{2}$ (principal values).

Step 1:

Given $\cos\left(\sin^{-1}\frac{2}{5}+\cos^{-1}x\right)=0$.

Since $\cos\theta = 0$ implies $\theta = \frac{\pi}{2}$ (within principal range), we have:

Step 2:

Recall the identity: $\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}$ for $t \in [-1,1]$. …

Concept and Intuition

The problem gives $\cos(\text{something}) = 0$. When does cosine equal zero? At $\frac{\pi}{2}$, $-\frac{\pi}{2}$, $\frac{3\pi}{2}$, etc. But here the "something" is a sum of two inverse trigonometric functions — each of which outputs an angle in a specific range. $\sin^{-1}\frac{2}{5}$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and $\cos^{-1}x$ lies in $[0, \pi]$. Their sum must fall in a range that makes cosine zero. The cleanest way is to recall a fundamental identity: for any $a$ in $[-1,1]$, $\sin^{-1}a + \cos^{-1}a = \frac{\pi}{2}$. That identity is the key — it tells us that if the two angles are inverses of the same number, their sum is exactly $\frac{\pi}{2}$, whose cosine is $0$. So the problem reduces to checking whether $\frac{2}{5}$ and $x$ are the same number.

Step-by-step solution

1. Set up the equation from the given condition We have $\cos\left(\sin^{-1}\frac{2}{5}+\cos^{-1}x\right)=0$. The general solution for $\cos\theta = 0$ is $\theta = \frac{\pi}{2} + n\pi$, where $n$ is an integer. So:

2. Restrict the possible values of $n$ using the ranges of the inverse functions
   • $\sin^{-1}\frac{2}{5}$ is a positive acute angle: $0 < \sin^{-1}\frac{2}{5} < \frac{\pi}{2}$.
   • $\cos^{-1}x$ lies in $[0, \pi]$. Therefore their sum lies strictly between $0$ and $\frac{3\pi}{2}$ (since the maximum is just under $\frac{\pi}{2} + \pi = \frac{3\pi}{2}$). The only value of $n$ that keeps $\frac{\pi}{2} + n\pi$ inside $(0, \frac{3\pi}{2})$ is $n=0$ (giving $\frac{\pi}{2}$). $n=1$ gives $\frac{3\pi}{2}$, which is not strictly less than $\frac{3\pi}{2}$ (and the sum cannot equal $\frac{3\pi}{2}$ exactly because $\sin^{-1}\frac{2}{5} < \frac{\pi}{2}$ and $\cos^{-1}x \le \pi$). So we must have: $\sin^{-1}\frac{2}{5}+\cos^{-1}x = \frac{\pi}{2}.$ …

Method: Solving $\cos(\text{sum of inverse-trig angles})=0$

An equation of the form $\cos(P+Q)=0$, where $P,Q$ are inverse-trig angles, is solved by turning the trig equation into a linear equation in those angles, then applying a known identity.

Steps

Step 1: Convert the outer trig equation to an angle equation.

$\cos\psi=0$ means $\psi=\tfrac{\pi}{2}+n\pi$ for integer $n$. So

Step 2: Restrict $n$ using the ranges of the inverse functions.

Bound $P$ and $Q$ by their principal ranges (e.g. $\sin^{-1}\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$, $\cos^{-1}\in[0,\pi]$). Their sum lies in a limited interval, which usually forces a single value of $n$ (here $n=0$, giving $P+Q=\tfrac{\pi}{2}$).

Step 3: Apply the relevant complementary identity. …

Common Mistakes

Mistake 1: Setting the inner sum to $0$ instead of $\tfrac{\pi}{2}$.

Why it's wrong: $\cos(\text{angle})=0$ means the angle is $\tfrac{\pi}{2}$ (plus multiples of $\pi$), not that the angle is $0$. Correct approach: $\sin^{-1}\tfrac25+\cos^{-1}x=\tfrac{\pi}{2}$.

Mistake 2: Concluding $x=-\tfrac25$ or some other value. …