Q.Show that $\cos\left(2\tan^{-1}\frac{1}{7}\right)=\sin\left(4\tan^{-1}\frac{1}{3}\right)$.

Concept: Trigonometric Simplification — Use double-angle identities and the tangent half-angle substitution.

Let $\alpha = \tan^{-1}\frac{1}{7}$ and $\beta = \tan^{-1}\frac{1}{3}$.

First, compute $\cos 2\alpha$. Using $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$:

Now compute $\sin 4\beta$. Use $\sin 4\beta = 2\sin 2\beta \cos 2\beta$, and $\sin 2\beta = \frac{2\tan\beta}{1+\tan^2\beta}$, $\cos 2\beta = \frac{1-\tan^2\beta}{1+\tan^2\beta}$:

$\sin 2\beta = \frac{2\cdot\frac{1}{3}}{1+\frac{1}{9}} = \frac{\frac{2}{3}}{\frac{10}{9}} = \frac{2}{3} \cdot \frac{9}{10} = \frac{6}{10} = \frac{3}{5},$ …

We need to show that two trigonometric expressions, each built from inverse tangents, are equal. The natural instinct is to let each inverse tangent be an angle — say $\alpha = \tan^{-1}\frac{1}{7}$ and $\beta = \tan^{-1}\frac{1}{3}$ — and then compute $\cos(2\alpha)$ and $\sin(4\beta)$ using known identities. Since $\tan\alpha$ and $\tan\beta$ are simple fractions, we can find $\cos(2\alpha)$ directly from $\tan\alpha$, and $\sin(4\beta)$ by first finding $\tan(2\beta)$ and then using the double-angle formula for sine. The whole thing reduces to checking whether both sides equal the same number.

1. Set up the angles.

   Let $\alpha = \tan^{-1}\frac{1}{7}$ and $\beta = \tan^{-1}\frac{1}{3}$.

   Then $\tan\alpha = \frac{1}{7}$ and $\tan\beta = \frac{1}{3}$.

2. Compute $\cos(2\alpha)$.

   There is a direct formula linking $\cos(2\theta)$ to $\tan\theta$:

This comes from $\cos(2\theta) = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta}$ and dividing numerator and denominator by $\cos^2\theta$.

So with $\tan\alpha = \frac{1}{7}$:

3. Compute $\sin(4\beta)$. We need $\sin(4\beta)$. A good path: first find $\tan(2\beta)$, then use $\sin(4\beta) = 2\sin(2\beta)\cos(2\beta)$, but we can also get $\sin(4\beta)$ directly from $\tan(2\beta)$ using another identity. Let’s find $\tan(2\beta)$ first:

Now we have $\tan(2\beta) = \frac{3}{4}$. This is a nice right-triangle ratio: opposite = 3, adjacent = 4, hypotenuse = 5. So:

Then $\sin(4\beta) = 2\sin(2\beta)\cos(2\beta) = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}$. …

Method: Comparing two multiple-angle expressions with the $t$-formulas

To prove two expressions built from inverse tangents are equal, convert each to a plain rational number using the tangent-only ("$t$") forms of the double-angle identities, then compare.

Steps

Step 1: Name each inverse and record its tangent.

Let $\alpha=\tan^{-1}p$, $\beta=\tan^{-1}q$, so $\tan\alpha=p$, $\tan\beta=q$.

Step 2: Use the $t$-formulas to get each side as a fraction.

$\cos 2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta},\qquad \sin 2\theta=\frac{2\tan\theta}{1+\tan^2\theta}.$ …

Common Mistakes

Mistake 1: Trying a single quadruple-angle formula for $\sin 4\beta$.

Why it's wrong: expanding $\sin 4\beta$ directly from $\tan\beta$ is long and error-prone. Correct approach: double twice — find $\tan 2\beta=\tfrac34$, read $\sin 2\beta=\tfrac35$, $\cos 2\beta=\tfrac45$, then $\sin 4\beta=2\cdot\tfrac35\cdot\tfrac45=\tfrac{24}{25}$.

Mistake 2: Using $\cos 2\theta=1-\tan^2\theta$ (dropping the denominator). …