Q12:

Find the values of each of the expressions: $\tan(\sin^{-1}(\frac{3}{5}) + \cot^{-1}(\frac{3}{2}))$

Step 1: Decomposition of the Expression

The given expression is $\tan(\sin^{-1}(\frac{3}{5}) + \cot^{-1}(\frac{3}{2}))$. To solve this, we must express both inverse trigonometric functions in terms of $\tan^{-1}(x)$.

Let $\alpha = \sin^{-1}(\frac{3}{5})$ and $\beta = \cot^{-1}(\frac{3}{2})$. The expression becomes $\tan(\alpha + \beta)$.

Step 2: Converting Inverse Functions to $\tan^{-1}$

For $\alpha = \sin^{-1}(\frac{3}{5})$:

• $\sin(\alpha) = \frac{3}{5}$ [Definition of inverse sine].
• Since $\sin(\alpha) = \frac{\text{Opposite}}{\text{Hypotenuse}}$, in a right-angled triangle, Opposite = 3, Hypotenuse = 5.
• Adjacent side = $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$ [By Pythagorean Theorem].
• $\tan(\alpha) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4}$.
• Therefore, $\alpha = \tan^{-1}(\frac{3}{4})$.

For $\beta = \cot^{-1}(\frac{3}{2})$:

• $\cot(\beta) = \frac{3}{2}$ [Definition of inverse cotangent].
• Using the identity $\tan^{-1}(x) = \cot^{-1}(\frac{1}{x})$ for $x > 0$:
• $\beta = \tan^{-1}(\frac{1}{3/2}) = \tan^{-1}(\frac{2}{3})$.

Step 3: Applying the Tangent Sum Identity

Now, the expression is $\tan(\tan^{-1}(\frac{3}{4}) + \tan^{-1}(\frac{2}{3}))$.

We use the identity: $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, provided $xy < 1$ [Summation property of inverse tangents].

Here, $x = \frac{3}{4}$ and $y = \frac{2}{3}$.

$xy = (\frac{3}{4}) \times (\frac{2}{3}) = \frac{6}{12} = \frac{1}{2}$. Since $\frac{1}{2} < 1$, the formula is valid.

Step 4: Final Algebraic Computation

Let $\theta = \tan^{-1}(\frac{3}{4}) + \tan^{-1}(\frac{2}{3})$.

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

$\theta = \tan^{-1}\left( \frac{\frac{9+8}{12}}{1 - \frac{6}{12}} \right)$

$\theta = \tan^{-1}\left( \frac{\frac{17}{12}}{\frac{6}{12}} \right) = \tan^{-1}\left( \frac{17}{6} \right)$

The original expression is $\tan(\theta) = \tan(\tan^{-1}(\frac{17}{6}))$.

By definition, $\tan(\tan^{-1}(z)) = z$ for all $z \in \mathbb{R}$.

Final Solution: \frac{17}{6}