Prove- sin - 1 1 x - cosec - 1 x - - x - 1 or x - - 1 cos - 1 1 x - - 1 x - - x - 1 or x - - 1 tan - 1 1 x - - 1 x - - x - 0

cos^ 1(1x)=θcosθ=1xsecθ =xsec^ 1x Similarly tan^ 11x=^ 1x ...

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Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

Prove: sin ...

Question

Prove:
${sin}^{- 1} (\frac{1}{x}) = c o s e c^{- 1} x, \forall x \geq 1 or x \leq - 1$
${cos}^{- 1} (\frac{1}{x}) = {sec}^{- 1} x, \forall x \geq 1 or x \leq - 1$
${tan}^{- 1} (\frac{1}{x}) = {cot}^{- 1} x, \forall x > 0$

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Solution

${cos}^{- 1} (\frac{1}{x}) = θ ⟹ cos θ = \frac{1}{x} ⟹ sec θ = x ⟹ {sec}^{- 1} x$
Similarly ${tan}^{- 1} \frac{1}{x} = {cot}^{- 1} x$

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\cot (\sin^{- 1} \frac{3}{4} + \sec^{- 1} \frac{4}{3})

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\sin (\tan^{- 1} x + \tan^{- 1} \frac{1}{x}) for x < 0

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Q. Write the value of tan⁻¹ x + tan⁻¹

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Q. Write the value of tan⁻¹x + tan⁻¹

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for x > 0.

Q. If x < 0, then tan^-1x + tan^-1

\frac{1}{x}

is equal to ____________________.

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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Prove:sin−1(1x)= cosec−1x,∀x≥1 or x≤−1cos−1(1x)=sec−1x,∀x≥1 or x≤−1tan−1(1x)=cot−1x,∀x>0

cos−1(1x)=θ⟹cosθ=1x⟹secθ=x⟹sec−1xSimilarly tan−11x=cot−1x

Prove:
${sin}^{- 1} (\frac{1}{x}) = c o s e c^{- 1} x, \forall x \geq 1 or x \leq - 1$
${cos}^{- 1} (\frac{1}{x}) = {sec}^{- 1} x, \forall x \geq 1 or x \leq - 1$
${tan}^{- 1} (\frac{1}{x}) = {cot}^{- 1} x, \forall x > 0$

${cos}^{- 1} (\frac{1}{x}) = θ ⟹ cos θ = \frac{1}{x} ⟹ sec θ = x ⟹ {sec}^{- 1} x$
Similarly ${tan}^{- 1} \frac{1}{x} = {cot}^{- 1} x$