Let us define the function as cos - 1 cos θ = θ and 2 tan - 1 x = tan - 1 2 x 1 - x 2 Statement I: if sin [ 2 cos - 1 { cot ( 2 tan - 1 x ) } ] = 0 , then x = ± 1 , ± 1 ± 2 Statement II: cot 2 tan -1 x = 1 - x 2 2 x

Statement I is true, Statement II is also true and Statement II is not the correct explanation of Statement I.

Statement I is true, Statement II is also true and Statement I is not the correct explanation of Statement II.

Statement I is true, Statement II is false.

Statement II is false, Statement I is true.

Let us define the function as cos - 1 cos θ = θ and 2 tan - 1 x = tan - 1 2 x 1 - x 2 Statement I: if sin [ 2 cos - 1 { cot ( 2 tan - 1 x ) } ] = 0 , then x = ± 1 , ± 1 ± 2 Statement II: cot 2 tan -1 x = 1 - x 2 2 x

Statement I is true, Statement II is also true and Statement II is not the correct explanation of Statement I.

Let f ( x ) = cos 2 tan - 1 sin cot - 1 1 - x x , 0 < x < 1 . Then:

( 1 - x ) 2 f ' ( x ) + 2 ( f ( x ) ) 2 = 0

( 1 + x ) 2 f ' ( x ) + 2 ( f ( x ) ) 2 = 0

( 1 - x ) 2 f ' ( x ) - 2 ( f ( x ) ) 2 = 0

( 1 + x ) 2 f ' ( x ) - 2 ( f ( x ) ) 2 = 0

HARD

Earn 100

Let us define the function as $\cos^{- 1} (\cos θ) = θ$ and $2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$

Statement I: if $\sin [2 \cos^{- 1} {\cot (2 \tan^{- 1} x)}] = 0,$ then $x = \pm 1, \pm (1 \pm \sqrt{2})$

Statement II: $\cot (2 \tan^{-1} x) = \frac{1 - x^{2}}{2 x}$

36.51% studentsanswered this correctly

Important Questions on Inverse Trigonometric Functions

EASY

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\sin^{- 1} (\frac{x}{5}) + \cos^{- 1} (\frac{4}{5}) = \frac{π}{2},

then the value of

x

is equal to

EASY

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\tan^{- 1} (\frac{1 - x}{1 + x}) - \frac{1}{2} \tan^{- 1} x = 0,

for

x > 0,

then

x =

MEDIUM

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\cot^{- 1} (\sqrt{\cos α}) - \tan^{- 1} (\sqrt{\cos α}) = x,

then

\sin x

is equal to

MEDIUM

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

The number of real roots of the equation

\tan^{- 1} \sqrt{x (x + 1)} + \sin^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{4}

is:

MEDIUM

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

The value of

\tan^{- 1} [\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}],

|x| < \frac{1}{2}, x \neq 0,

is equal to:

HARD

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

Which of the following values of

x

does not satisfy the equation

\sin (2 \cos^{- 1} (\cot (2 \tan^{- 1} x))) = 0

HARD

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\cos^{- 1} \sqrt{p} + \cos^{- 1} \sqrt{1 - p} + \cos^{- 1} \sqrt{1 - q} = \frac{3 π}{4},

then the value of

q

EASY

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\sin^{- 1} x - \cos^{- 1} x = \frac{π}{6}

, then

x

HARD

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

The value of

10 \tan (\cot^{- 1} 3 + \cot^{- 1} 7)

equal to

MEDIUM

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

The value of

\cos (\cos^{- 1} \frac{1}{5} + 2 \sin^{- 1} \frac{1}{5})

is equal to

EASY

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\cot (\sin^{- 1} x) = \cos (\tan^{- 1} \sqrt{3}),

then

x =

HARD

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

A value of

x

satisfying the equation

\sin [\cot^{- 1} (1 + x)] = \cos [\tan^{- 1} x],

is:

EASY

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

What is the value of

' x',

3 \tan^{- 1} {1 / (2 + \sqrt{3})} \tan^{- 1} x = \tan^{- 1} (1 / 3) ?

EASY

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\cot^{- 1} (α) = \cot^{- 1} 2 + \cot^{- 1} 8 + \cot^{- 1} 18 + \cot^{- 1} 32 + \dots .

upto

100

terms, then

α

is:

MEDIUM

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

Let

f (x) = \cos (2 \tan^{- 1} \sin (\cot^{- 1} \sqrt{\frac{1 - x}{x}})), 0 < x < 1

. Then:

MEDIUM

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation

\cos^{- 1} (x) - 2 \sin^{- 1} (x) = \cos^{- 1} (2 x)

is equal to

EASY

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\cos^{- 1} (\frac{5}{13}) - \sin^{- 1} (\frac{12}{13}) = \cos^{- 1} x

, then

x

is equal to

MEDIUM

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\sin^{- 1} x + \tan^{- 1} x = π / 2

, then

2 x^{2} + 1 =

EASY

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

4 {s i n}^{- 1} x + 6 {c o s}^{- 1} x = 3 π,

then

x =

MEDIUM

Mathematics>Trigonometry>Inverse Trigonometric Functions>Inverse Trigonometric Equations and Inequalities

\cos^{- 1} x - \cos^{- 1} \frac{y}{2} = α,

where

- 1 \leq x \leq 1, - 2 \leq y \leq 2, x \leq \frac{y}{2},

then for all

x, y, 4 x^{2} - 4 x y \cos α + y^{2}

is equal to :

Let us define the function as cos-1cosθ=θ and 2tan-1x=tan-12x1-x2 Statement I: if sin[2cos-1{cot(2tan-1x)}]=0, then x=±1,±1±2 Statement II: cot2tan-1x=1-x22x

Important Questions on Inverse Trigonometric Functions

Learn and Prepare for any exam you want !

Let us define the function as $\cos^{- 1} (\cos θ) = θ$ and $2 \tan^{- 1} x = \tan^{- 1} (\frac{2 x}{1 - x^{2}})$

Statement I: if $\sin [2 \cos^{- 1} {\cot (2 \tan^{- 1} x)}] = 0,$ then $x = \pm 1, \pm (1 \pm \sqrt{2})$

Statement II: $\cot (2 \tan^{-1} x) = \frac{1 - x^{2}}{2 x}$