Sin-1x tan-1xwhen x-0-1-

from above graph we can see that nbsp;sin^ 1x tan^ 1x in the interval (0, π/2] ...

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Standard XII

Mathematics

Plotting Graphs

sin-1x tan-1x...

Question

${sin}^{- 1} x$ ${tan}^{- 1} x$
when $x \in (0, 1]$

\geq

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\leq

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Solution

The correct option is C >

from above graph we can see that ${sin}^{- 1} x > {tan}^{- 1} x$ in the interval $(0, \frac{π}{2}]$

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Similar questions

{sin}^{- 1} x

{tan}^{- 1} x

when

x \in (0, 1]

{sin}^{- 1} x

{tan}^{- 1} x

when

x \in (0, 1]

Q. 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)

{s i n}^{- 1} (1 - x) - 2 {s i n}^{- 1} x = π / 2

.
(b) If

{s i n}^{- 1} x + {s i n}^{- 1} (1 - x) = {c o s}^{- 1} x

, then prove that x is equal to

0, 1 / 2

Q. If

f (x) = {tan}^{- 1} x - {cot}^{- 1} x,

g (x) = {sec}^{- 1} x - {cosec}^{- 1} x,

h (x) = {sin}^{- 1} x + {cos}^{- 1} x + {tan}^{- 1} x,

i (x) = {sin}^{- 1} x - {cos}^{- 1} x,

j (x) = \sqrt{x},

then

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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sin−1x tan−1x when x∈(0,1]

The correct option is C > from above graph we can see that sin−1x>tan−1x in the interval (0,π2]

${sin}^{- 1} x$ ${tan}^{- 1} x$
when $x \in (0, 1]$

The correct option is C >

from above graph we can see that ${sin}^{- 1} x > {tan}^{- 1} x$ in the interval $(0, \frac{π}{2}]$