Animated Solution for Mathematics - Inverse Trigonometric Functions: If ${\sin }^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\dots \right)+{\cos }^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\dots \right)=\frac{\pi }{2}$ for $0<|x|<\sqrt{2}$, then $x$ equals

A Journey Through Inverse Trigonometry

Welcome, fellow traveler of the JEE Advanced path. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of infinite series and inverse trigonometric functions.

It is designed to intimidate and make you panic. But as we peel back the layers, you will see that it is actually a beautifully orchestrated dance of symmetry and algebraic elegance.

The Lighthouse in the Storm

Imagine you are standing before the equation:

Your first instinct might be to try and evaluate these series directly or perhaps differentiate them. Stop. Take a breath.

In JEE Advanced, whenever you see ${\sin }^{-1}$ and ${\cos }^{-1}$ added together, your mind should immediately scream: Identity!

We know the fundamental identity: ${\sin }^{-1}(\theta )+{\cos }^{-1}(\theta )=\frac{\pi }{2}$. This is our lighthouse.

It tells us that for the sum to be exactly $\frac{\pi }{2}$, the arguments inside the inverse functions must be identical. We are not solving for the functions themselves; we are solving for the condition where the inputs are equal.

This realization transforms a terrifying calculus-looking problem into a simple algebraic equality: $A=B$.

The Infinite Geometric Progression

Now, let us look at the arguments. We have two infinite series. Let us analyze the first one: $S_1=x-\frac{x^2}{2}+\frac{x^3}{4}-\dots$.

Do you see the pattern? Each term is generated by multiplying the previous term by a constant factor. This is the definition of a Geometric Progression (G.P.).

The first term $a$ is $x$. To find the common ratio $r$, we take the second term and divide it by the first: $r=\frac{-x^2/2}{x}=-\frac{x}{2}$.

Using the sum formula for an infinite G.P., $S=\frac{a}{1-r}$, we get:

Now, look at the second series: $S_2=x^2-\frac{x^4}{2}+\frac{x^6}{4}-\dots$. This is also a G.P., but with a different starting point.

Here, $a=x^2$ and $r=-\frac{x^2}{2}$. Applying the same formula:

The Algebraic Dance

We have reduced the infinite complexity of the original equation to a simple rational equation:

This is where the magic happens. We know from the problem constraints that $0<|x|$, so $x eq 0$. This allows us to divide both sides by $2x$ without any guilt.

We are left with:

Cross-multiplying gives us $2+x^2=x(2+x)$, which expands to $2+x^2=2x+x^2$. Look at that!

The $x^2$ terms on both sides are identical. They cancel out, leaving us with the linear equation $2=2x$. Solving this is trivial: $x=1$.

In the heat of the exam, many students would stop here. But a true JEE Advanced aspirant knows that the final step is verification.

We must check our result against the constraints: $0<|x|<\sqrt{2}$. Since $|1|=1$, and $0<1<1.414$, our solution is valid.

We started with a complex, infinite expression and ended with a single integer. That is the power of recognizing patterns.

Never let the complexity of a problem intimidate you; look for the underlying structure, trust your identities, and let the algebra do the heavy lifting.