Q7:

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

Step 1: Analyzing the Right-Hand Side (RHS)

To prove the identity $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$, we first convert the individual inverse trigonometric terms on the RHS into $\tan^{-1}$ functions to utilize the addition formula: $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$.

Step 2: Conversion of Terms

• For $\sin^{-1} \frac{5}{13}$: Let $\alpha = \sin^{-1} \frac{5}{13}$. Thus, $\sin \alpha = \frac{5}{13}$. Using the Pythagorean identity $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{\frac{144}{169}} = \frac{12}{13}$. Therefore, $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{12/13} = \frac{5}{12}$. So, $\sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$. [Per the definition of trigonometric ratios in a right-angled triangle]
• For $\cos^{-1} \frac{3}{5}$: Let $\beta = \cos^{-1} \frac{3}{5}$. Thus, $\cos \beta = \frac{3}{5}$. Using the identity $\sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. Therefore, $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/5}{3/5} = \frac{4}{3}$. So, $\cos^{-1} \frac{3}{5} = \tan^{-1} \frac{4}{3}$.

Step 3: Applying the Addition Formula

Substitute the converted terms into the RHS expression:

$RHS = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$

Apply the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$ where $x = \frac{5}{12}$ and $y = \frac{4}{3}$:

Calculation of argument: $\frac{\frac{5}{12} + \frac{4}{3}}{1 - (\frac{5}{12} \times \frac{4}{3})}$

Numerator: $\frac{5 + 16}{12} = \frac{21}{12} = \frac{7}{4}$

Denominator: $1 - \frac{20}{36} = \frac{36 - 20}{36} = \frac{16}{36} = \frac{4}{9}$

Result of fraction: $\frac{7/4}{4/9} = \frac{7}{4} \times \frac{9}{4} = \frac{63}{16}$

Visual Representation of Triangle Conversions

Step 4: Final Conclusion

By equating the computed values, we obtain:

$\tan^{-1} \left( \frac{63}{16} \right) = \tan^{-1} \left( \frac{63}{16} \right)$

This confirms that $LHS = RHS$.

Final Solution: The identity $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$ is proven to be true.