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Basic Inverse Trigonometric Functions

Solve:- tan ...

Question

Solve:-
${tan}^{- 1} (\frac{6 x}{1 - 8 x^{2}})$

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Solution

${tan}^{- 1} \frac{6 x}{1 - {8 x}^{2}} = {tan}^{- 1} (\frac{2 x + 4 x}{1 - 2 x \times 4 x}) ({tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}) = {tan}^{- 1} 2 x + {tan}^{- 1} 4 x$

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Similar questions

Q. Solve for x:-

{tan}^{- 1} (\frac{6 x}{1 - 8 x^{2}}) = 0

{tan}^{- 1} (\frac{6 x}{1 - 8 x^{2}})

Solve : {tan}^{- 1} 4 x + {tan}^{- 1} 6 x = \frac{π}{4}

1 \leq \frac{3 x^{2} - 7 x + 8}{x^{2} + 1} \leq 2

Q. Solve the following inequalities.

\frac{1 - \sqrt{1 - 8 x^{2}}}{2 x} < 1

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