Q.If $2\tan^{-1}(\cos\theta)=\tan^{-1}(2\csc\theta)$, then show that $\theta=\frac{\pi}{4}$.

Concept: Inverse Tangent Identity

We use the formula $2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}$ for $x \in (-1,1)$.

Step 1: Let $x = \cos\theta$. Then the given equation becomes

Step 2: Since $\tan^{-1}$ is one-to-one, equate the arguments:

Step 3: Simplify using $1-\cos^2\theta = \sin^2\theta$ and $\csc\theta = \frac{1}{\sin\theta}$: …

The core of this problem is the inverse tangent double-angle identity. When you see $2\tan^{-1}(\text{something})$, your first instinct should be to rewrite it as a single $\tan^{-1}$ using:

Why does this work? Because if $\tan^{-1}x = \alpha$, then $\tan\alpha = x$, and $\tan(2\alpha) = \frac{2\tan\alpha}{1-\tan^2\alpha} = \frac{2x}{1-x^2}$. Taking $\tan^{-1}$ on both sides gives the identity. The condition $|x| < 1$ ensures the angle stays in the principal range $(-\pi/2, \pi/2)$, but here we'll check our final answer against the original equation.

Let's apply this to the given equation:

1. Apply the identity to the left-hand side. Let $x = \cos\theta$. Then:

2. Simplify the denominator using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$:

So the left side becomes:

3. Rewrite in terms of cosecant. Since $\csc\theta = \frac{1}{\sin\theta}$, we have:

But this isn't yet $2\csc\theta$. Let's keep it as is for now.

4. Equate the arguments of $\tan^{-1}$ on both sides. Since $\tan^{-1}$ is a one-to-one function on its principal range, if $\tan^{-1}A = \tan^{-1}B$, then $A = B$. So:

5. Cancel the common factor of 2 (assuming it's non-zero — we'll check later):

6. Rewrite $\csc\theta$ as $1/\sin\theta$:

7. Multiply both sides by $\sin^2\theta$ (valid as long as $\sin\theta \neq 0$; if $\sin\theta = 0$, the original equation has $\csc\theta$ undefined, so we can safely assume $\sin\theta \neq 0$):

8. Solve the trigonometric equation. $\cos\theta = \sin\theta$ implies:

The general solution is $\theta = \frac{\pi}{4} + n\pi$, where $n$ is an integer.

9. Check which solution fits the original equation. The original equation involves $\tan^{-1}(\cos\theta)$ and $\tan^{-1}(2\csc\theta)$. For $\theta = \frac{\pi}{4}$:
   • $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, so $2\tan^{-1}(\frac{1}{\sqrt{2}})$ is defined.
   • $\csc\frac{\pi}{4} = \sqrt{2}$, so $\tan^{-1}(2\sqrt{2})$ is defined.
   • Both sides are positive angles less than $\pi/2$, so the identity holds. …

Method: Solving an inverse-tangent equation with the doubling identity

For an equation such as $2\tan^{-1}(\text{expr})=\tan^{-1}(\text{expr})$, collapse the "$2$" into a single $\tan^{-1}$, then equate arguments because $\tan^{-1}$ is one-to-one.

Steps

Step 1: Apply the doubling identity to the $2\tan^{-1}$ side.

Step 2: Equate the arguments.

Once both sides are a single $\tan^{-1}(\cdot)$, injectivity of $\tan^{-1}$ lets you drop the inverses and set the insides equal.

Step 3: Simplify the trigonometric equation. …

Common Mistakes

Mistake 1: Equating the arguments before collapsing the $2\tan^{-1}$.

Why it's wrong: $2\tan^{-1}(\cos\theta)$ is not $\tan^{-1}(2\cos\theta)$, so you cannot set $2\cos\theta=2\csc\theta$. Correct approach: first apply $2\tan^{-1}x=\tan^{-1}\tfrac{2x}{1-x^2}$, then equate.

Mistake 2: Simplifying $1-\cos^2\theta$ incorrectly.

Why it's wrong: writing it as $\cos^2\theta$ or $\sin\theta$ breaks the algebra. Correct approach: $1-\cos^2\theta=\sin^2\theta$, so $\tfrac{2\cos\theta}{\sin^2\theta}=\tfrac{2}{\sin\theta}$ reduces to $\cos\theta=\sin\theta$. …