Question

The value of \( \cot(\csc^{-1}\frac{5}{3} + \tan^{-1}\frac{2}{3}) \) is

• A. 6/17
• B. 3/17
• C. 4/17
• D. 5/17

Hint:

When dealing with sums of different inverse trigonometric functions, the easiest approach is to convert all of them to \( \tan^{-1} \) using right-angled triangles. Then, apply the standard sum/difference formulas for \( \tan^{-1} \). Remember that \( \cot(\tan^{-1}x) = 1/x \).

Answer 1

The Correct Option is A

We need to evaluate \( \cot(A+B) \) where \( A = \csc^{-1}\frac{5}{3} \) and \( B = \tan^{-1}\frac{2}{3} \). 

First, convert \( A \) into a more common inverse trigonometric function like \( \tan^{-1} \). Let \( A = \csc^{-1}\frac{5}{3} \). This means \( \csc A = \frac{5}{3} \). Cosecant is Hypotenuse/Opposite. So, we have a right-angled triangle with Hypotenuse = 5 and Opposite = 3. 

By Pythagoras theorem, Adjacent = \( \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \). This is a standard 3-4-5 triangle. 

From this triangle, we can find \( \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4} \). 

So, \( A = \tan^{-1}\frac{3}{4} \). Now the expression becomes: \[ \cot\left(\tan^{-1}\frac{3}{4} + \tan^{-1}\frac{2}{3}\right) \] 

We use the sum formula for \( \tan^{-1}x + \tan^{-1}y \): \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \] \[ \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{2}{3} = \tan^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \cdot \frac{2}{3}}\right) = \tan^{-1}\left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right) = \tan^{-1}\left(\frac{17/12}{6/12}\right) = \tan^{-1}\left(\frac{17}{6}\right) \] So, the original expression is \( \cot\left(\tan^{-1}\frac{17}{6}\right) \). Since \( \cot(\tan^{-1}z) = 1/z \), we have: \[ \cot\left(\tan^{-1}\frac{17}{6}\right) = \frac{1}{17/6} = \frac{6}{17} \]