Q3:

Prove that $2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$

Step 1: Identifying the Left-Hand Side (LHS) Transformation

We begin with the expression: $2\sin^{-1} \left( \frac{3}{5} \right)$. To simplify this into a tangent form, we first express the $\sin^{-1}$ term as an angle $\theta$. Let:

$\sin^{-1} \left( \frac{3}{5} \right) = \theta \implies \sin \theta = \frac{3}{5}$ [Definition of Inverse Trigonometric Functions]

Step 2: Calculating Geometric Ratios

In a right-angled triangle where the opposite side $O = 3$ and the hypotenuse $H = 5$, we find the adjacent side $A$ using the Pythagorean theorem $A^2 + O^2 = H^2$:

$A = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$

Therefore, $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4}$.

Thus, $\sin^{-1} \left( \frac{3}{5} \right) = \tan^{-1} \left( \frac{3}{4} \right)$.

Step 3: Applying the Double Angle Formula

The original expression becomes $2\tan^{-1} \left( \frac{3}{4} \right)$. We utilize the standard identity for $2\tan^{-1} x$:

$2\tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 - x^2} \right)$ [Standard Identity for Inverse Tangent]

Substituting $x = \frac{3}{4}$:

$= \tan^{-1} \left( \frac{2(\frac{3}{4})}{1 - (\frac{3}{4})^2} \right)$

Step 4: Algebraic Simplification

Calculate the numerator and denominator:

• Numerator: $2 \times \frac{3}{4} = \frac{3}{2} = 1.5$
• Denominator: $1 - \frac{9}{16} = \frac{16-9}{16} = \frac{7}{16}$

Combining these into the tangent argument:

$\frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2} \times \frac{16}{7} = \frac{3 \times 8}{7} = \frac{24}{7}$

Step 5: Conclusion of Proof

Since the simplification yields $\tan^{-1} \left( \frac{24}{7} \right)$, which is exactly equal to the Right-Hand Side (RHS), the identity is proven.

Final Solution: $2\sin^{-1} \left( \frac{3}{5} \right) = \tan^{-1} \left( \frac{24}{7} \right)$