Inverse Trigonometric Functions Questions

MathematicsInverse Trigonometric Functions30 Questions Found

MathematicsSolving Inverse Trigonometric EquationsJEE Main 2002Moderate

$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

(A)

tan^{2} (\frac{α}{2})

(B)

cot^{2} (\frac{α}{2})

(C)

tan α

(D)

cot (\frac{α}{2})

Answer:A

MathematicsDomain and Range of Inverse Trigonometric FunctionsJEE Main 2002Easy

View in:English Hindi

The domain of $sin^{- 1} [lo g_{3} (x /3)]$ is

(A)

[1, 9]

(B)

[- 1, 9]

(C)

[- 9, 1]

(D)

[- 9, - 1]

Answer:A

MathematicsDomain and Range of Inverse Trigonometric FunctionsJEE Main 2003Easy

View in:English Hindi

The trigonometric equation $sin^{- 1} x = 2 sin^{- 1} a$ has a solution for

MathematicsProperties of Inverse Trigonometric FunctionsJEE Main 2005Easy

View in:English Hindi

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

(A)

2 \sin 2\alpha

(B)

(C)

4 \sin^2 \alpha

(D)

- 4 \sin^2 \alpha

Answer:C

MathematicsProperties of Inverse Trigonometric FunctionsJEE Main 2005Moderate

View in:English Hindi

Let $f : (- 1, 1) \to B$ , be a function defined by $f (x) = tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , then $f$ is both one - one and onto when $B$ is the interval

(A)

(0, π /2)

(B)

[0, π /2)

(C)

(- π /2, π /2)

(D)

(- π /2, π /2]

Answer:C

MathematicsSolving Inverse Trigonometric EquationsJEE Main 2007Easy

View in:English Hindi

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

(A)

(B)

(C)

(D)

Answer:D

MathematicsSolving Inverse Trigonometric EquationsJEE Main 2009Moderate

View in:English Hindi

Let $f (x) = (x + 1)^{2} - 1, x \geq - 1$ Statement-1 : The set ${x : f (x) = f^{- 1} (x)} = {0, - 1}$ Statement-2 : $f$ is a bijection.

(A)

Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.

(B)

Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1.

(C)

Statement-1 is true, Statement-2 is false.

(D)

Statement-1 is false, Statement-2 is true.

Answer:B

MathematicsProperties of Inverse Trigonometric FunctionsJEE Main 2013Moderate

View in:English Hindi

If $x, y, z$ are in A.P. and $tan^{- 1} x, tan^{- 1} y$ and $tan^{- 1} z$ are also in A.P., then

(A)

x = y = z

(B)

2x = 3y = 6z

(C)

6x = 3y = 2z

(D)

6x = 4y = 3z

Answer:A

MathematicsProperties of Inverse Trigonometric FunctionsJEE Main 2013Easy

View in:English Hindi

The value of $cot (cosec^{- 1} \frac{5}{3} + tan^{- 1} \frac{2}{3})$ is

(A)

6/17

(B)

3/17

(C)

4/17

(D)

5/17

Answer:A

MathematicsProperties of Inverse Trigonometric FunctionsJEE Main 2015Easy

View in:English Hindi

Let $tan^{- 1} y = tan^{- 1} x + tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $∣ x ∣ < \frac{1}{3}$ . Then a value of $y$ is :

(A)

\frac{3 x - x ^{3}}{1 + 3 x ^{2}}

(B)

\frac{3 x + x ^{3}}{1 + 3 x ^{2}}

(C)

\frac{3 x - x ^{3}}{1 - 3 x ^{2}}

(D)

\frac{3 x + x ^{3}}{1 - 3 x ^{2}}

Answer:C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cot−1(cosα​)−tan−1(cosα​)=x, then sinx=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The domain of sin−1[log3​(x/3)] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The trigonometric equation sin−1x=2sin−1a has a solution for

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If cos−1x−cos−12y​=α, then 4x2−4xycosα+y2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(−1,1)→B, be a function defined by f(x)=tan−1(1−x22x​), then f is both one - one and onto when B is the interval

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If sin−1(5x​)+cosec−1(45​)=2π​, then the values of x is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=(x+1)2−1,x≥−1 Statement-1 : The set {x:f(x)=f−1(x)}={0,−1} Statement-2 : f is a bijection.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x,y,z are in A.P. and tan−1x,tan−1y and tan−1z are also in A.P., then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of cot(cosec−135​+tan−132​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let tan−1y=tan−1x+tan−1(1−x22x​), where ∣x∣<3​1​. Then a value of y is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cot−1(cosα​)−tan−1(cosα​)=x, then sinx=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The domain of sin−1[log3​(x/3)] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The trigonometric equation sin−1x=2sin−1a has a solution for

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If cos−1x−cos−12y​=α, then 4x2−4xycosα+y2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(−1,1)→B, be a function defined by f(x)=tan−1(1−x22x​), then f is both one - one and onto when B is the interval

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If sin−1(5x​)+cosec−1(45​)=2π​, then the values of x is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=(x+1)2−1,x≥−1 Statement-1 : The set {x:f(x)=f−1(x)}={0,−1} Statement-2 : f is a bijection.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x,y,z are in A.P. and tan−1x,tan−1y and tan−1z are also in A.P., then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of cot(cosec−135​+tan−132​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let tan−1y=tan−1x+tan−1(1−x22x​), where ∣x∣<3​1​. Then a value of y is :

$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

The domain of $sin^{- 1} [lo g_{3} (x /3)]$ is

The trigonometric equation $sin^{- 1} x = 2 sin^{- 1} a$ has a solution for

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

Let $f : (- 1, 1) \to B$ , be a function defined by $f (x) = tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , then $f$ is both one - one and onto when $B$ is the interval

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

Let $f (x) = (x + 1)^{2} - 1, x \geq - 1$ Statement-1 : The set ${x : f (x) = f^{- 1} (x)} = {0, - 1}$ Statement-2 : $f$ is a bijection.

If $x, y, z$ are in A.P. and $tan^{- 1} x, tan^{- 1} y$ and $tan^{- 1} z$ are also in A.P., then

The value of $cot (cosec^{- 1} \frac{5}{3} + tan^{- 1} \frac{2}{3})$ is

Let $tan^{- 1} y = tan^{- 1} x + tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $∣ x ∣ < \frac{1}{3}$ . Then a value of $y$ is :

$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

The domain of $sin^{- 1} [lo g_{3} (x /3)]$ is

The trigonometric equation $sin^{- 1} x = 2 sin^{- 1} a$ has a solution for

If $cos^{- 1} x - cos^{- 1} \frac{y}{2} = α$ , then $4 x^{2} - 4 x y cos α + y^{2}$ is equal to

Let $f : (- 1, 1) \to B$ , be a function defined by $f (x) = tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , then $f$ is both one - one and onto when $B$ is the interval

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

Let $f (x) = (x + 1)^{2} - 1, x \geq - 1$ Statement-1 : The set ${x : f (x) = f^{- 1} (x)} = {0, - 1}$ Statement-2 : $f$ is a bijection.

If $x, y, z$ are in A.P. and $tan^{- 1} x, tan^{- 1} y$ and $tan^{- 1} z$ are also in A.P., then

The value of $cot (cosec^{- 1} \frac{5}{3} + tan^{- 1} \frac{2}{3})$ is

Let $tan^{- 1} y = tan^{- 1} x + tan^{- 1} (\frac{2 x}{1 - x ^{2}})$ , where $∣ x ∣ < \frac{1}{3}$ . Then a value of $y$ is :