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

Concept: Inverse Cosine Addition — convert each inverse trigonometric function into a tangent form, then use the tangent addition formula.

Step 1: Let $\alpha = \sin^{-1}\frac{5}{13}$. Then $\sin\alpha = \frac{5}{13}$, so $\cos\alpha = \frac{12}{13}$ and $\tan\alpha = \frac{5}{12}$.

Step 2: Let $\beta = \cos^{-1}\frac{3}{5}$. Then $\cos\beta = \frac{3}{5}$, so $\sin\beta = \frac{4}{5}$ and $\tan\beta = \frac{4}{3}$.

Step 3: We want $\tan(\alpha + \beta)$: …

We need to show that the sum of an inverse sine and an inverse cosine equals a specific inverse tangent. The direct approach — taking sine or cosine of both sides — gets messy because the left side is a sum of two different inverse functions. A cleaner path is to express each term as an inverse tangent, because tangent addition is straightforward.

Why tangent?

If $A = \sin^{-1} \frac{5}{13}$ and $B = \cos^{-1} \frac{3}{5}$, then $A+B$ is some angle. We can find $\tan(A+B)$ using known values of $\tan A$ and $\tan B$. If that equals $\frac{63}{16}$, and we also check that $A+B$ lies in the correct range for $\tan^{-1}$, the identity holds.

Let’s do it step by step.

1. Find $\tan A$ where $A = \sin^{-1} \frac{5}{13}$

   If $\sin A = \frac{5}{13}$, then by the Pythagorean identity, $\cos A = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$. Since $\sin^{-1}$ gives an angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and $\frac{5}{13} > 0$, $A$ is in the first quadrant, so $\cos A$ is positive.

   Hence $\tan A = \frac{\sin A}{\cos A} = \frac{5/13}{12/13} = \frac{5}{12}$.

2. Find $\tan B$ where $B = \cos^{-1} \frac{3}{5}$

   If $\cos B = \frac{3}{5}$, then $\sin B = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The range of $\cos^{-1}$ is $[0, \pi]$, and $\frac{3}{5} > 0$ puts $B$ in the first quadrant, so $\sin B$ is positive.

   Thus $\tan B = \frac{\sin B}{\cos B} = \frac{4/5}{3/5} = \frac{4}{3}$.

3. Apply the tangent addition formula

   For any angles $A$ and $B$ (where $\cos(A+B) \neq 0$):

Substitute $\tan A = \frac{5}{12}$ and $\tan B = \frac{4}{3}$:

Compute numerator: $\frac{5}{12} + \frac{4}{3} = \frac{5}{12} + \frac{16}{12} = \frac{21}{12} = \frac{7}{4}$.

Compute denominator: $1 - \frac{5 \cdot 4}{12 \cdot 3} = 1 - \frac{20}{36} = 1 - \frac{5}{9} = \frac{4}{9}$.

So:

4. Check the range …

Method: Combining mixed inverse functions by converting to tangent

When a sum mixes $\sin^{-1}$ and $\cos^{-1}$ and the target is a $\tan^{-1}$, convert every term to its tangent, use the tangent addition formula, then range-check.

Steps

Step 1: Turn each inverse term into a tangent value.

If $A=\sin^{-1}s$ then $\tan A=\dfrac{s}{\sqrt{1-s^2}}$; if $B=\cos^{-1}c$ then $\tan B=\dfrac{\sqrt{1-c^2}}{c}$. (Build the right triangle to see these.)

Step 2: Apply the tangent addition formula.

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.$ …

Common Mistakes

Mistake 1: Adding the arguments as if $\sin^{-1}s+\cos^{-1}c=\sin^{-1}(s+c)$ or similar.

Why it's wrong: inverse-trig values are angles; you cannot add their arguments. Correct approach: convert each to a tangent, then use $\tan(A+B)=\tfrac{\tan A+\tan B}{1-\tan A\tan B}$.

Mistake 2: Slipping in $\tan A$ or $\tan B$.

Why it's wrong: from $\sin^{-1}\tfrac{5}{13}$ the tangent is $\tfrac{5}{12}$ (not $\tfrac{5}{13}$), and from $\cos^{-1}\tfrac35$ it is $\tfrac43$. Correct approach: build each triangle first, then $\tan(A+B)=\tfrac{63}{16}$. …