Q.Find the value of the following: $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$

Concept: Inverse Tangent Identity – When adding inverse cosines and sines, converting to inverse tangents often simplifies the sum because $\tan^{-1} a + \tan^{-1} b$ has a clean closed form.

Let $\alpha = \cos^{-1} \frac{12}{13}$ and $\beta = \sin^{-1} \frac{3}{5}$.

Step 1: Find $\tan \alpha$ and $\tan \beta$.

For $\alpha$, $\cos \alpha = \frac{12}{13}$, so $\sin \alpha = \frac{5}{13}$ (since $\alpha \in [0,\pi]$ and sine is positive). Hence $\tan \alpha = \frac{5}{12}$.

For $\beta$, $\sin \beta = \frac{3}{5}$, so $\cos \beta = \frac{4}{5}$ (since $\beta \in [-\frac{\pi}{2},\frac{\pi}{2}]$ and cosine is positive). Hence $\tan \beta = \frac{3}{4}$.

Step 2: Use the sum formula for inverse tangent:

provided $ab < 1$. Here $a = \frac{5}{12}$, $b = \frac{3}{4}$, and $ab = \frac{5}{16} < 1$. So

$\tan(\alpha + \beta) = \frac{\frac{5}{12} + \frac{3}{4}}{1 - \frac{5}{16}} = \frac{\frac{5}{12} + \frac{9}{12}}{\frac{11}{16}} = \frac{\frac{14}{12}}{\frac{11}{16}} = \frac{14}{12} \cdot \frac{16}{11} = \frac{56}{33}.$ …

The core idea here is that when you have inverse trigonometric functions of rational numbers, it's often easier to work with their tangents. Why? Because the tangent of an angle is a ratio of sine and cosine, and the tangent addition formula is straightforward: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Once we find $\tan(A+B)$, we can recover the angle's sine using a right triangle.

Let’s set:

• $A = \cos^{-1} \frac{12}{13}$, so $\cos A = \frac{12}{13}$.
• $B = \sin^{-1} \frac{3}{5}$, so $\sin B = \frac{3}{5}$.

We want to show $A + B = \sin^{-1} \frac{56}{65}$.

1. Find $\tan A$ from $\cos A$.

   Since $\cos A = \frac{12}{13}$, we can imagine a right triangle where adjacent = 12, hypotenuse = 13. By Pythagoras, opposite = $\sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$.

   So $\sin A = \frac{5}{13}$ (positive because $A$ is in $[0, \pi]$ for $\cos^{-1}$, and $\cos A > 0$ means $A$ is in $[0, \frac{\pi}{2}]$, so sine is positive).

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

2. Find $\tan B$ from $\sin B$.

   $\sin B = \frac{3}{5}$, so opposite = 3, hypotenuse = 5. Adjacent = $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.

   Since $B = \sin^{-1} \frac{3}{5}$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and the sine is positive, $B$ is in $[0, \frac{\pi}{2}]$, so $\cos B = \frac{4}{5}$ (positive).

   Thus $\tan B = \frac{3}{4}$.

3. Apply the tangent addition formula.

   Let $C = A + B$. Then:

Compute numerator: $\frac{5}{12} + \frac{3}{4} = \frac{5}{12} + \frac{9}{12} = \frac{14}{12} = \frac{7}{6}$.

Denominator: $1 - \frac{5 \cdot 3}{12 \cdot 4} = 1 - \frac{15}{48} = 1 - \frac{5}{16} = \frac{11}{16}$.

So $\tan C = \frac{7/6}{11/16} = \frac{7}{6} \times \frac{16}{11} = \frac{112}{66} = \frac{56}{33}$.

4. Find $\sin C$ from $\tan C$. We have $\tan C = \frac{56}{33}$. In a right triangle, opposite = 56, adjacent = 33. Hypotenuse = $\sqrt{56^2 + 33^2} = \sqrt{3136 + 1089} = \sqrt{4225} = 65$. Therefore $\sin C = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{56}{65}$. …

Method: Combining mixed inverse-sine and inverse-cosine terms

Use this to evaluate a sum such as $\cos^{-1}x + \sin^{-1}y$ as a single inverse function.

Steps

Step 1: Convert every term to a common trig ratio.

Turn each inverse term into an angle and find its tangent from a right triangle. This lets one addition formula handle both.

Step 2: Apply the tangent addition formula.

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

Common Mistakes

Mistake 1: Assuming $\sin(A+B) = \sin A + \sin B$.

Why it's wrong: sine is not additive; you must use the addition formula. Correct approach: convert to tangents and use $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$, then recover the sine.

Mistake 2: Skipping the quadrant check on $A+B$. …