For alpha, beta, gamma neq 0. If -1alpha + -1beta + -1gamma = pi and (alpha + beta + gamma)(alpha - gamma + beta) = 3 alphabeta then gamma equal to

Let -1alpha = A, -1beta = B, -1gamma = C. Given A + B + C = pi. Since A = alpha, B = beta, C = gamma, alpha, beta, gamma act like the side lengths of a triangle divided by 2R by Sine rule. However, directly dealing with the relation: Divide by 2alphabeta: By Cosine Rule, C = (1/2). Since C = -1gamma, we know C = gamma. C = sqrt(1 - gamma2) = (1/2). Since C is an angle of a triangle (or sum equals pi and elements are positive limits), gamma = C > 0. The expression (alpha + beta + gamma)(alpha + beta - gamma) = 3alphabeta perfectly mirrors the Cosine Rule standard form giving C = 1/2. Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Trigonometric Functions

For alpha, beta, gamma neq 0. If -1alpha + -1beta + -1gamma = pi and (alpha + beta + gamma)(alpha - gamma + beta) = 3 alphabeta then gamma equal to

Let -1alpha = A, -1beta = B, -1gamma = C. Given A + B + C = pi. Since A = alpha, B = beta, C = gamma, alpha, beta, gamma act like the side lengths of a triangle divided by 2R by Sine rule. However, directly dealing with the relation: Divide by 2alphabeta: By Cosine Rule, C = (1/2). Since C = -1gamma, we know C = gamma. C = sqrt(1 - gamma2) = (1/2). Since C is an angle of a triangle (or sum equals pi and elements are positive limits), gamma = C > 0. The expression (alpha + beta + gamma)(alpha + b

For alpha, beta, gamma neq 0. If -1alpha + -1beta + -1gamma = pi and (alpha + beta + gamma)(alpha - gamma + beta) = 3 alphabeta then gamma equal to

Let -1alpha = A, -1beta = B, -1gamma = C. Given A + B + C = pi. Since A = alpha, B = beta, C = gamma, alpha, beta, gamma act like the side lengths of a triangle divided by 2R by Sine rule. However, directly dealing with the relation: Divide by 2alphabeta: By Cosine Rule, C = (1/2). Since C = -1gamma, we know C = gamma. C = sqrt(1 - gamma2) = (1/2). Since C is an angle of a triangle (or sum equals pi and elements are positive limits), gamma = C > 0. The expression (alpha + beta + gamma)(alpha + beta - gamma) = 3alphabeta perfectly mirrors the Cosine Rule standard form giving C = 1/2. Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Trigonometric Functions

For alpha, beta, gamma neq 0. If -1alpha + -1beta + -1gamma = pi and (alpha + beta + gamma)(alpha - gamma + beta) = 3 alphabeta then gamma equal to

Let -1alpha = A, -1beta = B, -1gamma = C. Given A + B + C = pi. Since A = alpha, B = beta, C = gamma, alpha, beta, gamma act like the side lengths of a triangle divided by 2R by Sine rule. However, directly dealing with the relation: Divide by 2alphabeta: By Cosine Rule, C = (1/2). Since C = -1gamma, we know C = gamma. C = sqrt(1 - gamma2) = (1/2). Since C is an angle of a triangle (or sum equals pi and elements are positive limits), gamma = C > 0. The expression (alpha + beta + gamma)(alpha + b

For alpha, beta, gamma neq 0. If -1alpha + -1beta + -1gamma = pi and (alpha + beta + gamma)(alpha

Q10 jee_main_2026_24_january_morning Half Angle and Compound Angle Formulas

If

x = (5)/(12)

for some

x in ( π, (3π)/(2) )

, then

7x( (13x)/(2) + (13x)/(2) ) + 7x( (13x)/(2) - (13x)/(2) )

is equal to

A. $4\sqrt(26)$ $\frac{4}{\sqrt{26}}$
B. $6\sqrt(26)$ $\frac{6}{\sqrt{26}}$
C. $1\sqrt(13)$ $\frac{1}{\sqrt{13}}$
D. $5\sqrt(13)$ $\frac{5}{\sqrt{13}}$

Solution

Related Formula

(A + B) = A B + A B

(A - B) = A B - A B

x = 2 ²((x)/(2)) - 1 = 1 - 2 ²((x)/(2))

Core Logic

Given $x = (5)/(12)$ $\cot x = \frac{5}{12}$ and $x in (π, 3π/2)$ $x \in (\pi, 3\pi/2)$ (3rd quadrant), $x = (-5)/(13)$ $\cos x = \frac{-5}{13}$. Find $(x/2)$ $\cos(x/2)$ and $(x/2)$ $\sin(x/2)$. Since $π < x < (3π)/(2) \Rightarrow (π)/(2) < (x)/(2) < (3π)/(4)$ (2nd quadrant, so $> 0$ $\sin > 0$, $< 0$ $\cos < 0$).

Step 1: Half Angle Values

x = (-5)/(13) = 2 ²((x)/(2)) - 1 \Rightarrow 2 ²((x)/(2)) = (8)/(13) \Rightarrow ((x)/(2)) = - 2\sqrt(13)

-1 + 2 ²((x)/(2)) = (5)/(13) \Rightarrow ((x)/(2)) = 3\sqrt(13)

Step 2: Expression Simplification

The expression is:

E = 7x (13x)/(2) + 7x (13x)/(2) + 7x (13x)/(2) - 7x (13x)/(2)

Regrouping:

E = ( 7x (13x)/(2) - 7x (13x)/(2)) + ( 7x (13x)/(2) + 7x (13x)/(2))

Using identities $(A-B)$ $\sin(A-B)$ and $(A-B)$ $\cos(A-B)$:

E = (7x - (13x)/(2)) + (7x - (13x)/(2))

E = ((x)/(2)) + ((x)/(2))

Step 3: Final Calculation

E = 3\sqrt(13) + (- 2\sqrt(13)) = 1\sqrt(13)

Pattern Recognition

Expanding and rearranging mixed sine/cosine terms often collapses large coefficients ( $7x, 13x/2$ $7x, 13x/2$) into their simple difference ( $x/2$ $x/2$).

Chapter Mix

Class 11 Maths: Trigonometric Functions

Q25 jee_main_2026_24_january_evening General Solutions of Trigonometric Equations

The number of elements in the set

x in [0, 180°] : (x + 100°) = (x + 50°) x (x - 50°)

is

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

Componendo and Dividendo: (a)/(b) = (c)/(d) (a+b)/(a-b) = (c+d)/(c-d)

A B \pm A B = (A \pm B)

Core Logic

Rewrite the given equation by dividing $x$ $\tan x$ to the LHS:

(x + 100°) x = (x + 50°) (x - 50°)

Convert all terms to sines and cosines:

(x + 100°) x (x + 100°) x = (x + 50°) (x - 50°) (x + 50°) (x - 50°)

Step 1: Componendo and Dividendo

Apply Componendo and Dividendo on both sides:

(x + 100°) x + (x + 100°) x (x + 100°) x - (x + 100°) x = (x + 50°) (x - 50°) + (x + 50°) (x - 50°) (x + 50°) (x - 50°) - (x + 50°) (x - 50°)

Simplify numerators and denominators using addition formulas:

(2x + 100°) (100°) = (100°)- (2x)

Cross multiply:

- (2x + 100°) (2x) = (100°) (100°)

Step 2: Trigonometric Simplification

Multiply by 2:

-2 (2x + 100°) (2x) = 2 (100°) (100°)

Use product-to-sum formulas:

- ( (4x + 100°) + (100°)) = (200°)

(4x + 100°) + (100°) + (200°) = 0

Combine the constant sines:

(100°) + (200°) = 2 (150°) (-50°) = 2 ((1)/(2)) (50°) = (50°)

So,

(4x + 100°) = - (50°)

Step 3: Solving for x

Convert $- (50°)$ $-\cos(50^{\circ})$ into sine:

- (50°) = - (40°) = (-40°)

Therefore:

(4x + 100°) = (-40°)

The general solution is:

4x + 100° = n(180°) + (-1)ⁿ (-40°)

4x = n(180°) + (-1)ⁿ⁺¹(40°) - 100°

x = n(45°) + (-1)ⁿ⁺¹(10°) - 25°

Step 4: Finding Specific Roots

Substitute integers for $n$ $n$ to find roots in the interval $x in [0, 180°]$ $x \in [0, 180^{\circ}]$: For $n = 1$ $n = 1$: $x = 45° + (1)(10°) - 25° = 30°$ For $n = 2$ $n = 2$: $x = 90° + (-1)(10°) - 25° = 55°$ For $n = 3$ $n = 3$: $x = 135° + (1)(10°) - 25° = 120°$ For $n = 4$ $n = 4$: $x = 180° + (-1)(10°) - 25° = 145°$ For $n = 5$ $n = 5$: $x = 225° + (1)(10°) - 25° = 210°$ (Out of bounds)

There are exactly $4$ $4$ solutions in $[0, 180°]$ $[0, 180^{\circ}]$.

Pattern Recognition

Fractional equality of sine and cosine products heavily points to Componendo and Dividendo. Pushing $x$ $\tan x$ to the left immediately creates the standard format ready to collapse back into $(A \pm B)$ $\sin(A \pm B)$ and $(A \pm B)$ $\cos(A \pm B)$.

Chapter Mix

Class 11 Maths: Trigonometric Equations

Q4 jee_main_2026_28_january_morning Trigonometric Identities

If

( (A-B))/( A) + ( ² C)/( ² A) = 1

,

A, B, C in (0, (π)/(2))

, then

A. $A, C, B are in G.P.$
B. $A, B, C are in G.P.$
C. $A, C, B are in A.P.$
D. $A, B, C are in A.P.$

Solution

Related Formula

(A-B) = ( A - B)/(1 + A B)

(1)/( ² θ) = 1 + ² θ

Core Logic

Given equation:

( A - B)/((1 + A B) A) + (1 + ² A)/(1 + ² C) = 1

Let $A = x$ $\tan A = x$, $B = y$ $\tan B = y$, $C = z$ $\tan C = z$. Substitute these into the equation:

(x - y)/(x(1 + xy)) + ((1 + (1)/(x²)))/((1 + (1)/(z²))) = 1

(x - y)/(x(1 + xy)) + ((x² + 1)z²)/(x²(z² + 1)) = 1

Step 1: Algebraic Simplification

Multiply through to clear denominators:

x(x - y)(z² + 1) + z² (1 + x²)(1 + xy) = x²(1 + xy)(z² + 1)

Expanding both sides:

(x² - xy)(z² + 1) + z²(1 + xy + x² + x³ y) = x²(z² + 1 + xyz² + xy)

x²z² + x² - xyz² - xy + z² + xyz² + x²z² + x³yz² = x²z² + x² + x³yz² + x³y

Cancel common terms on both sides to isolate the relation between $x, y, z$ $x, y, z$.

Step 2: Conclusion

After fully expanding and rearranging, we obtain:

z² (1 + x²) = xy (1 + x²)

Since $A in (0, (π)/(2))$ $A \in \left(0, \frac{\pi}{2}\right)$, $x \neq 0$ $x \neq 0$ and $1 + x² \neq 0$ $1 + x^2 \neq 0$, we can divide by $(1+x²)$ $(1+x^2)$: $z² = xy$ $z^2 = xy$ Substituting back our tangent terms:

² C = A \cdot B

This is the condition for a Geometric Progression.

Step 3: Final Answer

$A$ $\tan A$, $C$ $\tan C$, $B$ $\tan B$ are in G.P.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions Class 11 Mathematics: Sequences and Series

Q23 jee_main_2026_28_january_morning Properties of Inverse Trigonometric Functions

If

k = ((π)/(4) +(1)/(2) ⁻¹((2)/(3))) + ((1)/(2) ⁻¹((2)/(3)))

then the number of solutions of the equation

⁻¹(kx - 1) = ⁻¹x - ⁻¹x

is ____.

Numerical Answer. Answer: 1 to 1

Solution

Core Logic

First, evaluate $k$ $k$. Let $θ = (1)/(2) ⁻¹((2)/(3))$ . This implies $(2θ) = (2)/(3)$ $\sin(2\theta) = \frac{2}{3}$. Since $⁻¹ x + ⁻¹ x = (π)/(2)$ , we have:

⁻¹((2)/(3)) = (π)/(2) - ⁻¹((2)/(3))

(1)/(2) ⁻¹((2)/(3)) = (π)/(4) - (1)/(2) ⁻¹((2)/(3)) = (π)/(4) - θ

Substitute this back into $k$ $k$:

k = ((π)/(4) + (π)/(4) - θ) + (θ)

k = ((π)/(2) - θ) + (θ)

k = θ + θ

Step 1: Simplify k

k = ( θ)/( θ) + ( θ)/( θ) = ( ²θ + ²θ)/( θ θ) = (1)/( θ θ)

Multiply by $2/2$ $2/2$:

k = (2)/(2 θ θ) = (2)/( (2θ))

Since $(2θ) = (2)/(3)$ $\sin(2\theta) = \frac{2}{3}$:

k = (2)/(2/3) = 3

Step 2: Solve the Equation

Now solve $⁻¹(3x - 1) = ⁻¹x - ⁻¹x$ . We know $⁻¹x = (π)/(2) - ⁻¹x$ .

⁻¹(3x - 1) = ⁻¹x - ((π)/(2) - ⁻¹x)

⁻¹(3x - 1) = 2 ⁻¹x - (π)/(2)

⁻¹(3x - 1) = -((π)/(2) - 2 ⁻¹x)

Take sine of both sides:

3x - 1 = (-((π)/(2) - 2 ⁻¹x))

3x - 1 = - (2 ⁻¹x)

Step 3: Finding Roots

Let $⁻¹x = α$ $\sin^{-1}x = \alpha$, so $x = α$ $x = \sin\alpha$.

3x - 1 = - (2α) = -(1 - 2 ²α) = 2x² - 1

$2x² - 3x = 0$ $2x^2 - 3x = 0$ $x(2x - 3) = 0$ $x(2x - 3) = 0$ $x = 0$ $x = 0$ or $x = (3)/(2)$ $x = \frac{3}{2}$. Since domain of $⁻¹$ $\sin^{-1}$ is $[-1, 1]$ $[-1, 1]$, $x = 3/2$ $x = 3/2$ is rejected.

Now check $x=0$ $x=0$ in original equation: LHS: $⁻¹(-1) = -π/2$ $\sin^{-1}(-1) = -\pi/2$ RHS: $⁻¹(0) - ⁻¹(0) = 0 - π/2 = -π/2$ Both sides match, so $x=0$ $x=0$ is a valid solution. Wait, the official solution says $x=0$ $x=0$ is rejected and number of solutions is 1? No, the snippet says "x=0, 3/2 (rejected)" which implies 3/2 is rejected. Then it says "No. of solution = 1". So $x=0$ $x=0$ is indeed the 1 solution.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions Class 11 Mathematics: Trigonometric Functions

Q6 jee_main_2026_28_january_evening Transformation of Inverse Trig Expressions

Considering the principal values of inverse trigonometric functions, the value of the expression

(2 ⁻¹( 2\sqrt(13))-2 ⁻¹( 3\sqrt(10)))

is equal to:

A. $-(33)/(56)$ $-\frac{33}{56}$
B. $(33)/(56)$ $\frac{33}{56}$
C. $(16)/(63)$ $\frac{16}{63}$
D. $-(16)/(63)$ $-\frac{16}{63}$

Solution

Related Formula

(A - B) = ( A - B)/(1 + A B)

2θ = (2 θ)/(1 - ²θ)

Core Logic

Let $⁻¹ 2\sqrt(13) = θ$ $\sin^{-1}\frac{2}{\sqrt{13}} = \theta$ and $⁻¹ 3\sqrt(10) = φ$ $\cos^{-1}\frac{3}{\sqrt{10}} = \phi$. Then $θ = 2\sqrt(13)$ $\sin\theta = \frac{2}{\sqrt{13}}$ $\Rightarrow$ $\Rightarrow$ $θ = (2)/(3)$ $\tan\theta = \frac{2}{3}$. And $φ = 3\sqrt(10)$ $\cos\phi = \frac{3}{\sqrt{10}}$ $\Rightarrow$ $\Rightarrow$ $φ = (1)/(3)$ $\tan\phi = \frac{1}{3}$.

Execution

Calculate $2θ$ $\tan 2\theta$:

2θ = (2(2/3))/(1 - (4/9)) = (4/3)/(5/9) = (12)/(5)

Calculate $2φ$ $\tan 2\phi$:

2φ = (2(1/3))/(1 - (1/9)) = (2/3)/(8/9) = (6)/(8) = (3)/(4)

Now, substitute into the $(2θ - 2φ)$ $\tan(2\theta - 2\phi)$ identity:

(2θ - 2φ) = ((12)/(5) - (3)/(4))/(1 + ((12)/(5))((3)/(4)))

= ((48 - 15)/(20))/(1 + (36)/(20)) = ((33)/(20))/((56)/(20)) = (33)/(56)

Pattern Recognition

For compound inverse trigonometric expressions involving coefficients, map them entirely to their $\tan$ equivalents early to avoid cumbersome algebraic radicals.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions

For α, β, γ ≠ 0$\alpha, \beta, \gamma \neq 0$. If ⁻¹α + ⁻¹β + ⁻¹γ = π$\sin^{-1}\alpha + \sin^{-1}\beta + \sin^{-1}\gamma = \pi$ and (α + β + γ)(α - γ + β) = 3 αβ$(\alpha + \beta + \gamma)(\alpha - \gamma + \beta) = 3 \alpha\beta$ then γ$\gamma$ equal to

Solution & Explanation

Core Logic

Step 1: Simplify Algebraic Relation

Step 2: Triangle Identification

Step 3: Final Solution

Pattern Recognition

Chapter Mix

Reference Study Guides

More Inverse Trigonometric Functions Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Half Angle Values

Step 2: Expression Simplification

Step 3: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Componendo and Dividendo

Step 2: Trigonometric Simplification

Step 3: Solving for x

Step 4: Finding Specific Roots

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Algebraic Simplification

Step 2: Conclusion

Step 3: Final Answer

Chapter Mix

Solution

Core Logic

Step 1: Simplify k

Step 2: Solve the Equation

Step 3: Finding Roots

Chapter Mix

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

More Inverse Trigonometric Functions Questions — jee_main_2024_31_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

For $α, β, γ \neq 0$ $\alpha, \beta, \gamma \neq 0$. If $⁻¹α + ⁻¹β + ⁻¹γ = π$ and $(α + β + γ)(α - γ + β) = 3 αβ$ then $γ$ $\gamma$ equal to