Q.$\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$ Prove that

Concept: Inverse Tangent Identity — We use the sum formula for inverse tangents:

$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x + y}{1 - xy}$, valid when $xy < 1$.

Step 1: Convert the right side to $\tan^{-1}$ form.

Let $\alpha = \sin^{-1} \frac{5}{13}$, so $\tan \alpha = \frac{5}{12}$.

Let $\beta = \cos^{-1} \frac{3}{5}$, so $\tan \beta = \frac{4}{3}$.

Step 2: Apply the sum formula: …

We need to prove:

The left side is a single inverse tangent. The right side is a sum of two different inverse trigonometric functions. The natural strategy is to convert everything to $\tan^{-1}$ so we can use the addition formula for inverse tangents.

Why this works: If we can show that the tangent of the right-hand side equals $\frac{63}{16}$, and that both sides lie in the same quadrant (so the inverse tangent gives the same principal value), then the identity holds.

1. Convert $\sin^{-1} \frac{5}{13}$ to $\tan^{-1}$

   Let $\alpha = \sin^{-1} \frac{5}{13}$. Then $\sin \alpha = \frac{5}{13}$.

   Using the Pythagorean identity: $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.

   (Since $\sin^{-1}$ gives an angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and $\frac{5}{13} > 0$, $\alpha$ is in the first quadrant, so $\cos \alpha$ is positive.)

   Therefore, $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{12/13} = \frac{5}{12}$.

   So $\alpha = \tan^{-1} \frac{5}{12}$.

2. Convert $\cos^{-1} \frac{3}{5}$ to $\tan^{-1}$

   Let $\beta = \cos^{-1} \frac{3}{5}$. Then $\cos \beta = \frac{3}{5}$.

   Then $\sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

   (Again, $\cos^{-1}$ gives an angle in $[0, \pi]$, and $\frac{3}{5} > 0$, so $\beta$ is in the first quadrant, $\sin \beta$ positive.)

   Hence $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/5}{3/5} = \frac{4}{3}$.

   So $\beta = \tan^{-1} \frac{4}{3}$.

3. Now the right-hand side becomes:

4. Apply the inverse tangent addition formula

   For $xy < 1$: $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}$

   For $xy > 1$ and $x, y > 0$: $\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \frac{x+y}{1-xy}$ (since the sum exceeds $\frac{\pi}{2}$).

   Here $x = \frac{5}{12}$, $y = \frac{4}{3}$. Compute $xy = \frac{5}{12} \cdot \frac{4}{3} = \frac{20}{36} = \frac{5}{9} < 1$.

   So we use the first case:

5. Simplify the fraction Numerator: $\frac{5}{12} + \frac{4}{3} = \frac{5}{12} + \frac{16}{12} = \frac{21}{12} = \frac{7}{4}$. Denominator: $1 - \frac{5}{9} = \frac{4}{9}$. So the argument becomes: …

Method: Proving a $\tan^{-1}$ identity by converting every term to inverse tangent

Use this for proofs like $\tan^{-1}c = \sin^{-1}a + \cos^{-1}b$.

Steps

Step 1: Rewrite each inverse sine/cosine as an inverse tangent.

From $\sin\alpha = a$, build a triangle to get $\tan\alpha$; from $\cos\beta = b$, do the same. Now every term is a $\tan^{-1}$.

Step 2: Apply the addition formula, checking the $xy$ condition.

$\tan^{-1}p + \tan^{-1}q = \tan^{-1}\!\left(\frac{p+q}{1-pq}\right), \quad pq < 1.$ …

Common Mistakes

Mistake 1: Reading the wrong tangent from $\cos^{-1}\frac{3}{5}$.

Why it's wrong: $\cos\beta = \frac{3}{5}$ gives $\sin\beta = \frac{4}{5}$, so $\tan\beta = \frac{4}{3}$, not $\frac{3}{4}$. Correct approach: get the missing side by Pythagoras and form $\tan = \frac{\sin}{\cos}$.

Mistake 2: Quoting the addition formula without checking $pq < 1$. …