Question

Find the value of tan inverse of root 3 minus cot inverse of minus root 3

Answer 1

Hey there!

Interesting Question.

Let's detail, and uncover the whole process for this question; asking for a inverse in the tangential and cotangential functions, respectively.

Presume the first variable to cover for a function of inverse in tangential value as "a", subsequently the non-inverse or real tangential function to be represented as variable "b"; that is:

$\boxed{\bf{a = tan^{- 1} (b) \qquad Or, \: \: \: a = arctan(b) \implies tan(a) = b, \: \: Range \: \: for; \: \: \: -\dfrac{\pi}{2} < a < \dfrac{\pi}{2}}}$

$\boxed{\bf{\therefore \quad x = arctan(\sqrt{3}) \implies tan(x) = \sqrt{3}, \: \: Range \: \: for; \: \: \: -\dfrac{\pi}{2} < x < \dfrac{\pi}{2}}}$

So, solving for a function of the tangent:

$\bf{tan(x) = \sqrt{3}, \quad - \dfrac{\pi}{2} < x < \dfrac{\pi}{2}}$

$\bf{General \: \: Solutions \: \: for \: \: this \: \: tangent: \\ \\ \\ x = \dfrac{\pi}{3} + \pi n \\ \\ x = \dfrac{\pi}{3} + \pi n}$

The solutions satisfying the range criteria for this query will be as "n = 0"; Since, it's covering in that particular range. So, final value for the first tangential function is:

$\bf{For \: \: tangent: \quad x = \dfrac{\pi}{3}}$

$\bf{\therefore \quad \Bigg(\dfrac{\pi}{3} \Bigg) - arccot \Big(- \sqrt{3} \Big)}$

Similarly;

$\boxed{\bf{a = cot^{- 1} (b) \qquad Or, \: \: \: a = arccot(b) \implies cot(a) = b, \: \: Range \: \: for; \: \: \: 0 < a < \pi}}$

$\boxed{\bf{\therefore \quad x = arccot \Big(-\sqrt{3} \Big) \implies cot(x) = -\sqrt{3}, \: \: Range \: \: for; \: \: \: 0 < x < \pi}}$

Again; Solving for a function of the cotangent:

$\bf{cot(x) = -\sqrt{3}, \quad 0 < x < \pi}$

$\bf{General \: \: Solutions \: \: for \: \: this \: \: cotangent: \\ \\ \\ x = \dfrac{5\pi}{6} + \pi n \\ \\ x = \dfrac{5\pi}{6} + \pi n}$

Therefore;

$\bf{= \: \Bigg(\dfrac{\pi}{3} \Bigg) - \Bigg(\dfrac{5\pi}{6} \Bigg)}$

$\bf{= \: \dfrac{2\pi}{6} - \dfrac{5\pi}{6}}$

$\bf{= \: \dfrac{2\pi - 5\pi}{6}}$

$\bf{= \: \dfrac{-3\pi}{6}}$

$\boxed{\bf{\underline{\therefore \quad Final \: \: Answer \: \: is: \: \: tan^{-1}(\sqrt{3}) - cot^{-1}(-\sqrt{3}) = -\dfrac{\pi}{2}}}}$

Which is the required solution for these types of queries.

Hope it helps you and clears your doubt for finding the values of inverse functions!!!!