Solving Inverse Trigonometric Equations

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JEE Main 2009

LEVELJEE Main

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

JEE Main 2007

LEVELJEE Main

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

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

(A)

tan^{2} (\frac{α}{2})

(B)

cot^{2} (\frac{α}{2})

(C)

tan α

(D)

cot (\frac{α}{2})

JEE Advanced 2004

LEVELJEE Main

The value of $x$ for which $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$ is

The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = π /2$ is

Match List I with List II:

List-I

(P)

[\frac{1}{y ^{2}} (\frac{c o s ( t a n ^{- 1} y ) + y s i n ( t a n ^{- 1} y )}{c o t ( s i n ^{- 1} y ) + t a n ( s i n ^{- 1} y )})^{2} + y^{4}]^{1/2}

takes value

(Q)

cos x + cos y + cos z = 0 = sin x + sin y + sin z

then possible value of

cos \frac{x - y}{2}

(R)

cos (\frac{π}{4} - x) cos 2 x + sin x sin 2 x sec x = cos x sin 2 x sec x + cos (\frac{π}{4} + x) cos 2 x

then possible value of

sec x

(S)

cot (sin^{- 1} 1 - x^{2}) = sin (tan^{- 1} (x 6))

x \neq = 0

, then possible value of

x

List-II

(1)

\frac{1}{2} \frac{5}{3}

(2)

2

(3)

\frac{1}{2}

(4)

JEE Advanced 2007

LEVELJEE Main

Let $(x, y)$ be such that $sin^{- 1} (a x) + cos^{- 1} (y) + cos^{- 1} (b x y) = \frac{π}{2}$ . Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

(P)

a = 1

and

b = 0

, then

(x, y)

(Q)

a = 1

and

b = 1

, then

(x, y)

(R)

a = 1

and

b = 2

, then

(x, y)

(S)

a = 2

and

b = 2

, then

(x, y)

List-II

(1)

lies on the circle

x^{2} + y^{2} = 1

(2)

lies on

(x^{2} - 1) (y^{2} - 1) = 0

(3)

lies on

y = x

(4)

lies on

(4 x^{2} - 1) (y^{2} - 1) = 0

JEE Main 2019 (12 January Shift 1)

LEVELJEE Main

Considering only the principal values of inverse functions, the set $A = {x \geq 0 : tan^{- 1} (2 x) + tan^{- 1} (3 x) = π /4}$

(A)

is an empty set

(B)

Contains more than two elements

(C)

Contains two elements

(D)

is a singleton

JEE Main 2019 (10 January)

LEVELJEE Main

The value of $cot (\sum_{n = 1}^{19} cot^{- 1} (1 + \sum_{p = 1}^{n} 2 p))$ is :

(A)

\frac{22}{23}

(B)

\frac{23}{22}

(C)

\frac{21}{19}

(D)

\frac{19}{21}

JEE Main 11 Jan 2019 (Evening)

LEVELJEE Main

All $x$ satisfying the inequality $(cot^{- 1} x)^{2} - 7 (cot^{- 1} x) + 10 > 0$ , lie in the interval:

(A)

(- \infty, cot 5) \cup (cot 4, cot 2)

(B)

(cot 5, cot 4)

(C)

(- \infty, cot 5) \cup (cot 2, \infty)

(D)

(cot 2, \infty)

JEE Main 2021 (27 Aug Shift 2)

LEVELJEE Main

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f (x) = tan^{- 1} (sin x + cos x)$ in $[0, \frac{π}{2}]$ , Then the value of $tan (M - m)$ is equal to:

(A)

2 + 3

(B)

2 - 3

(C)

3 + 22

(D)

3 - 22

JEE Main 2021 (20 July Shift 1)

LEVELJEE Main

The number of real roots of the equation $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$ is :

JEE Main 2021 (February)

LEVELJEE Main

If $0 < a, b < 1$ , and $tan^{- 1} a + tan^{- 1} b = \frac{π}{4}$ , then the value of $(a + b) - \frac{( a ^{2} + b ^{2} )}{2} + \frac{( a ^{3} + b ^{3} )}{3} - \frac{( a ^{4} + b ^{4} )}{4} + \dots$ is :

(A)

lo g_{e} 2

(B)

lo g_{e} (\frac{e}{2})

(C)

e

(D)

e^{2} - 1

JEE Main 2021 (16 March Shift 1)

LEVELJEE Main

Let $S_{k} = \sum_{r = 1}^{k} tan^{- 1} (\frac{6 ^{r}}{2 ^{2 r + 1} + 3 ^{2 r + 1}})$ . Then $lim_{k \to \infty} S_{k}$ equal to :

(A)

tan^{- 1} (\frac{3}{2})

(B)

\frac{π}{2}

(C)

cot^{- 1} (\frac{3}{2})

(D)

tan^{- 1} (3)

JEE Main 2021 (16 March Shift 2)

LEVELJEE Main

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy $sin^{- 1} (\frac{3 x}{5}) + sin^{- 1} (\frac{4 x}{5}) = sin^{- 1} x$ is equal to:

JEE Main 2021 (17 March Shift 1)

LEVELJEE Main

The sum of possible values of $x$ for $tan^{- 1} (x + 1) + cot^{- 1} (\frac{1}{x - 1}) = tan^{- 1} (\frac{8}{31})$ is

(A)

- \frac{32}{4}

(B)

- \frac{31}{4}

(C)

- \frac{30}{4}

(D)

- \frac{33}{4}

JEE Main 2021 (17 March Shift 2)

LEVELJEE Main

The number of solutions of the equation $sin^{- 1} [x^{2} + \frac{1}{3}] + cos^{- 1} [x^{2} - \frac{2}{3}] = x^{2}$ for $x \in [- 1, 1]$ , and $[x]$ denotes the greatest integer less than or equal to $x$ , is :

JEE Main 2022 (24 June Shift 2)

LEVELJEE Main

Let $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$ . Then a value of $2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})$ is

(A)

\frac{π}{4}

(B)

\frac{π}{3}

(C)

\frac{π}{2}

(D)

\frac{π}{6}

JEE Main 2022 (28 July Shift 1)

LEVELJEE Main

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 :

(A)

(B)

(C)

\frac{1}{2}

(D)

- \frac{1}{2}

JEE Main 2023 (13 April Shift 1)

LEVELJEE Main

If $S = {x \in R : sin^{- 1} (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}}$ then $\sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π))$ is equal to _________.

JEE Main 2023 (31 January Shift 1)

LEVELJEE Main

If $sin^{- 1} \frac{α}{17} + cos^{- 1} \frac{4}{5} - tan^{- 1} \frac{77}{36} = 0, 0 < α < 13$ , then $sin^{- 1} (sin α) + cos^{- 1} (cos α)$ is equal to

(A)

π

π

JEE Main 2023 (31 January Shift 2)

LEVELJEE Main

Let $(a, b) \subset (0, 2 π)$ be the largest interval for which $sin^{- 1} (sin θ) - cos^{- 1} (sin θ) > 0, θ \in (0, 2 π)$ , holds. If $α x^{2} + β x + sin^{- 1} (x^{2} - 6 x + 10) + cos^{- 1} (x^{2} - 6 x + 10) = 0$ and $α - β = b - a$ , then $α$ is equal to :

(A)

\frac{π}{48}

(B)

\frac{π}{16}

(C)

\frac{π}{8}

(D)

\frac{π}{12}

JEE Main 2023 (01 February Shift 1)

LEVELJEE Main

Let $S$ be the set of all solutions of the equation $cos^{- 1} (2 x) - 2 cos^{- 1} (1 - x^{2}) = π, x \in [- \frac{1}{2}, \frac{1}{2}]$ . Then $\sum_{x \in S} 2 sin^{- 1} (x^{2} - 1)$ is equal to

(A)

(B)

\frac{- 2 π}{3}

(C)

π - sin^{- 1} (\frac{3}{4})

(D)

π - 2 sin^{- 1} (\frac{3}{4})

JEE Main 2024 (06 Apr Shift 1)

LEVELJEE Main

For $n \in N$ , if $cot^{- 1} 3 + cot^{- 1} 4 + cot^{- 1} 5 + cot^{- 1} n = \frac{π}{4}$ , then $n$ is equal to

JEE Main 2024 (27 January Shift 2)

LEVELJEE Main

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$ 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 \\ \textbf{Statement-1 :} The set {x:f(x)=f−1(x)}={0,−1} \\ \textbf{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 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; } 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 value of x for which sin(cot−1(1+x))=cos(tan−1x) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of real solutions of tan−1x(x+1)​+sin−1x2+x+1​=π/2 is

Match List I with List II:

List-I

List-II

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Considering only the principal values of inverse functions, the set A={x≥0:tan−1(2x)+tan−1(3x)=π/4}

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of cot(∑n=119​cot−1(1+∑p=1n​2p)) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } All x satisfying the inequality (cot−1x)2−7(cot−1x)+10>0, lie in the interval:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let M and m respectively be the maximum and minimum values of the function f(x)=tan−1(sinx+cosx) in [0,2π​], Then the value of tan(M−m) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of real roots of the equation tan−1x(x+1)​+sin−1x2+x+1​=4π​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 0<a,b<1, and tan−1a+tan−1b=4π​, then the value of (a+b)−2(a2+b2)​+3(a3+b3)​−4(a4+b4)​+… is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Sk​=∑r=1k​tan−1(22r+1+32r+16r​). Then limk→∞​Sk​ equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of possible values of x for tan−1(x+1)+cot−1(x−11​)=tan−1(318​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of the equation sin−1[x2+31​]+cos−1[x2−32​]=x2 for x∈[−1,1], and [x] denotes the greatest integer less than or equal to x, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let x∗y=x2+y3 and (x∗1)∗1=x∗(1∗1). Then a value of 2sin−1(x4+x2+2x4+x2−2​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos−1(x)−2sin−1(x)=cos−1(2x) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If S={x∈R:sin−1(x2+2x+2​x+1​)−sin−1(x2+1​x​)=4π​} then ∑x∈S​(sin((x2+x+5)2π​)−cos((x2+x+5)π)) is equal to _________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If sin−117α​+cos−154​−tan−13677​=0,0<α<13, then sin−1(sinα)+cos−1(cosα) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the set of all solutions of the equation cos−1(2x)−2cos−1(1−x2​)=π,x∈[−21​,21​]. Then ∑x∈S​2sin−1(x2−1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For n∈N, if cot−13+cot−14+cot−15+cot−1n=4π​, then n is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying tan−1(x)+tan−1(2x)=4π​ is :

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

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

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

The value of $x$ for which $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$ is

The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = π /2$ is

Considering only the principal values of inverse functions, the set $A = {x \geq 0 : tan^{- 1} (2 x) + tan^{- 1} (3 x) = π /4}$

The value of $cot (\sum_{n = 1}^{19} cot^{- 1} (1 + \sum_{p = 1}^{n} 2 p))$ is :

All $x$ satisfying the inequality $(cot^{- 1} x)^{2} - 7 (cot^{- 1} x) + 10 > 0$ , lie in the interval:

Let $M$ and $m$ respectively be the maximum and minimum values of the function $f (x) = tan^{- 1} (sin x + cos x)$ in $[0, \frac{π}{2}]$ , Then the value of $tan (M - m)$ is equal to:

The number of real roots of the equation $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$ is :

If $0 < a, b < 1$ , and $tan^{- 1} a + tan^{- 1} b = \frac{π}{4}$ , then the value of $(a + b) - \frac{( a ^{2} + b ^{2} )}{2} + \frac{( a ^{3} + b ^{3} )}{3} - \frac{( a ^{4} + b ^{4} )}{4} + \dots$ is :

Let $S_{k} = \sum_{r = 1}^{k} tan^{- 1} (\frac{6 ^{r}}{2 ^{2 r + 1} + 3 ^{2 r + 1}})$ . Then $lim_{k \to \infty} S_{k}$ equal to :

The sum of possible values of $x$ for $tan^{- 1} (x + 1) + cot^{- 1} (\frac{1}{x - 1}) = tan^{- 1} (\frac{8}{31})$ is

The number of solutions of the equation $sin^{- 1} [x^{2} + \frac{1}{3}] + cos^{- 1} [x^{2} - \frac{2}{3}] = x^{2}$ for $x \in [- 1, 1]$ , and $[x]$ denotes the greatest integer less than or equal to $x$ , is :

Let $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$ . Then a value of $2 sin^{- 1} (\frac{x ^{4} + x ^{2} - 2}{x ^{4} + x ^{2} + 2})$ is

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 :

If $S = {x \in R : sin^{- 1} (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}}$ then $\sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π))$ is equal to _________.

If $sin^{- 1} \frac{α}{17} + cos^{- 1} \frac{4}{5} - tan^{- 1} \frac{77}{36} = 0, 0 < α < 13$ , then $sin^{- 1} (sin α) + cos^{- 1} (cos α)$ is equal to

Let $S$ be the set of all solutions of the equation $cos^{- 1} (2 x) - 2 cos^{- 1} (1 - x^{2}) = π, x \in [- \frac{1}{2}, \frac{1}{2}]$ . Then $\sum_{x \in S} 2 sin^{- 1} (x^{2} - 1)$ is equal to

For $n \in N$ , if $cot^{- 1} 3 + cot^{- 1} 4 + cot^{- 1} 5 + cot^{- 1} n = \frac{π}{4}$ , then $n$ is equal to

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$ is :