Question

The value of $17cot(cosec^{-1}\frac{5}{3}+tan^{-1}\frac{2}{3})$ is _____.

Answer 1

6

Answer 2

Let the given expression be $E$. We have $E = 17cot(cosec^{-1}\frac{5}{3}+tan^{-1}\frac{2}{3})$.

First, we evaluate the term inside the cotangent function. Let $A = cosec^{-1}\frac{5}{3}$. This implies $cosec A = \frac{5}{3}$. Since $cosec A = \frac{1}{sin A}$, we have $sin A = \frac{3}{5}$. We can convert $sin^{-1}\frac{3}{5}$ to $tan^{-1}$. Consider a right triangle with the opposite side 3 and the hypotenuse 5. The adjacent side is $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$. Thus, $tan A = \frac{opposite}{adjacent} = \frac{3}{4}$. So, $cosec^{-1}\frac{5}{3} = sin^{-1}\frac{3}{5} = tan^{-1}\frac{3}{4}$.

Now, the expression inside the cotangent function is $tan^{-1}\frac{3}{4} + tan^{-1}\frac{2}{3}$. We use the formula for the sum of two inverse tangents: $tan^{-1}x + tan^{-1}y = tan^{-1}\left(\frac{x+y}{1-xy}\right)$, provided $xy < 1$. Here $x = \frac{3}{4}$ and $y = \frac{2}{3}$. The product $xy = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$. Since $xy = \frac{1}{2} < 1$, we can use the formula: $tan^{-1}\frac{3}{4} + tan^{-1}\frac{2}{3} = tan^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times\frac{2}{3}}\right) = tan^{-1}\left(\frac{\frac{9+8}{12}}{1-\frac{6}{12}}\right) = tan^{-1}\left(\frac{\frac{17}{12}}{\frac{1}{2}}\right)$. Simplifying the fraction: $\frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{12} \times 2 = \frac{17}{6}$. So, $cosec^{-1}\frac{5}{3}+tan^{-1}\frac{2}{3} = tan^{-1}\frac{17}{6}$.

The expression $E$ becomes $17cot(tan^{-1}\frac{17}{6})$. We know that $cot(\theta) = \frac{1}{tan(\theta)}$. So, $cot(tan^{-1}x) = \frac{1}{tan(tan^{-1}x)} = \frac{1}{x}$ for $x \neq 0$. Here $x = \frac{17}{6}$, which is not zero. $cot(tan^{-1}\frac{17}{6}) = \frac{1}{\frac{17}{6}} = \frac{6}{17}$.

Finally, we substitute this value back into the expression for $E$: $E = 17 \times cot(tan^{-1}\frac{17}{6}) = 17 \times \frac{6}{17}$. $E = 6$.