Q.Find the principal value of the following: $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\right)$

Concept: Inverse Trigonometric Graphs — we evaluate the expression by converting each inverse function into an angle in a right triangle, then using the tangent addition formula.

Let $\alpha = \sin^{-1}\frac{3}{5}$. Then $\sin\alpha = \frac{3}{5}$, so the opposite side is $3$ and hypotenuse $5$, giving adjacent $4$ (by Pythagoras). Hence $\tan\alpha = \frac{3}{4}$.

Let $\beta = \cot^{-1}\frac{3}{2}$. Then $\cot\beta = \frac{3}{2}$, so $\tan\beta = \frac{2}{3}$.

Now use the tangent addition formula: …

We need to evaluate $\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\right)$. The expression inside the tangent is a sum of two angles — one from an inverse sine, the other from an inverse cotangent. The direct approach is to find the tangent of each angle separately, then use the formula for $\tan(A+B)$.

Let’s set:

We want $\tan(A+B)$.

1. Find $\tan A$ from $\sin A = \frac{3}{5}$ Since $A = \sin^{-1}\frac{3}{5}$, we know $\sin A = \frac{3}{5}$ and $A$ lies in the principal range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Here $\frac{3}{5}>0$, so $A$ is in the first quadrant. Draw a right triangle: opposite side = 3, hypotenuse = 5. By Pythagoras, adjacent side = $\sqrt{5^2 - 3^2} = \sqrt{25-9} = \sqrt{16} = 4$. Hence $\cos A = \frac{4}{5}$ and

2. Find $\tan B$ from $\cot B = \frac{3}{2}$

   $B = \cot^{-1}\frac{3}{2}$ means $\cot B = \frac{3}{2}$. The principal range of $\cot^{-1}$ is $(0, \pi)$, and since $\frac{3}{2}>0$, $B$ is in the first quadrant.

   $\cot B = \frac{\text{adjacent}}{\text{opposite}} = \frac{3}{2}$, so in a right triangle: adjacent = 3, opposite = 2. Hypotenuse = $\sqrt{3^2 + 2^2} = \sqrt{9+4} = \sqrt{13}$.

   Then $\tan B = \frac{1}{\cot B} = \frac{2}{3}$.

   Tip

   You don’t need the hypotenuse for $\tan B$ — just reciprocate $\cot B$. But the triangle helps if you later need $\sin B$ or $\cos B$.

3. Apply the tangent addition formula

Substitute $\tan A = \frac{3}{4}$ and $\tan B = \frac{2}{3}$:

4. Simplify numerator and denominator …

Method: Tangent of a sum of inverse-trigonometric angles

Use this for any expression of the form $\tan\!\left(\text{(inverse-trig)} + \text{(inverse-trig)}\right)$.

Steps

Step 1: Name each inverse term as an angle.

Set $A$ and $B$ equal to the two inverse expressions, so each becomes an ordinary angle whose one known ratio you can read directly.

Step 2: Find $\tan A$ and $\tan B$ using a right triangle. …

Common Mistakes

Mistake 1: Using $\tan(\cot^{-1}\frac{3}{2}) = \frac{3}{2}$.

Why it's wrong: if $\cot\beta = \frac{3}{2}$, then $\tan\beta$ is the reciprocal, $\frac{2}{3}$, not $\frac{3}{2}$. Correct approach: reciprocate the cotangent to get the tangent before substituting.

Mistake 2: Using a plus sign in the denominator of the addition formula. …