The angle between the lines represented by the equation 4 x 2-24 xy -11 y 2-0 areA- tan -13-4- tan -1-3-4B- tan -11-3- tan -1-1-3C- tan -14-3- tan -1-4-3D- tan -11-2- tan -1-1-2

The correct option is C tan θ = ± 2√(h^2 ab)/a + b = ± 2√(144 44)/15 = ±4/3. ⇒θ = tan^ 1(±4/3). ...

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Standard XII

Mathematics

Inclination of a Line

The angle bet...

Question

The angle between the lines represented by the equation $4 x^{2} - 24 x y + 11 y^{2} = 0$ are

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Solution

The correct option is C

$t a n θ = \pm 2 \frac{\sqrt{h^{2} - a b}}{a + b} = \pm 2 \frac{\sqrt{144 - 44}}{15} = \pm \frac{4}{3}$ .

$\Rightarrow θ = t a n^{- 1} (\pm \frac{4}{3})$ .

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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}

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(a)

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Q. Show that

{tan}^{- 1} 1 / 2 + {tan}^{- 1} 2 / 11 + {tan}^{- 1} 4 / 3 = π / 2

Q. Prove

{tan}^{- 1} \frac{1}{2} + {tan}^{- 1} \frac{2}{11} = {tan}^{- 1} \frac{3}{4}

Q. Prove that

{tan}^{- 1} \frac{1}{2} + {tan}^{- 1} \frac{2}{11} = {tan}^{- 1} \frac{3}{4}

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

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The angle between the lines represented by the equation 4x2−24xy+11y2=0are

The correct option is C tanθ=±2√h2−aba+b=±2√144−4415=±43. ⇒θ=tan−1(±43).

The angle between the lines represented by the equation $4 x^{2} - 24 x y + 11 y^{2} = 0$ are

The correct option is C

$t a n θ = \pm 2 \frac{\sqrt{h^{2} - a b}}{a + b} = \pm 2 \frac{\sqrt{144 - 44}}{15} = \pm \frac{4}{3}$ .

$\Rightarrow θ = t a n^{- 1} (\pm \frac{4}{3})$ .