Find m if tan - 1 1-x2 - 1-1x- m-4- x - - 1-

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

(a)

{t a n}^{- 1} \frac{1}{4} + 2 {t a n}^{- 1} \frac{1}{5} + {t a n}^{- 1} \frac{1}{6} + {t a n}^{- 1} \frac{1}{x} = \frac{π}{4}

(b)

{t a n}^{- 1} (x - 1) + {t a n}^{- 1} x + {t a n}^{- 1} (x + 1) = {t a n}^{- 1} 3 x

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Find m if tan−11√x2−1+sec−1x=mπ4,|x|>1.

tan−11√x2−1+sec−1x=mπ4Let x=secθ.tan−11√sec2θ−1+sec−1(secθ)=mπ4tan−11tanθ+θ=mπ4tan−1(cotθ)+θ=mπ4tan−1(tan{π2−θ})+θ=mπ4.π2−θ+θ=mπ4we get mπ4=π2∴m=2

Find m if ${tan}^{- 1} \frac{1}{\sqrt{x^{2} - 1}} + {sec}^{- 1} x = \frac{m π}{4}, | x | > 1$ .