If alpha, -(pi/2) alpha (pi/2) is the solution of 4+5=1, then the value of is

Year	2026	2025	2024	Total
Questions	15	18	10	43

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

²θ - ²θ = 1

If $a θ + b θ = c$ $a\cos\theta + b\sin\theta = c$, dividing by $θ$ $\cos\theta$ transforms the equation into a quadratic in terms of $θ$ $\tan\theta$ and $θ$ $\sec\theta$.

Core Logic

Given the equation:

4 θ + 5 θ = 1

Divide the entire equation by $θ$ $\cos\theta$:

4 + 5 θ = θ

Square both sides to convert the $θ$ $\sec\theta$ into a $θ$ $\tan\theta$ expression:

(4 + 5 θ)² = ²θ

16 + 25 ²θ + 40 θ = 1 + ²θ

Rearranging into a standard quadratic equation in terms of $θ$ $\tan\theta$:

24 ²θ + 40 θ + 15 = 0

Step 1: Apply Quadratic Formula

Solve for $θ$ $\tan\theta$ using the quadratic formula:

θ = -40 \pm \sqrt(1600 - 4(24)(15))2(24)

θ = -40 \pm \sqrt(1600 - 1440)48

θ = -40 \pm \sqrt(160)48

θ = -40 \pm 4\sqrt(10)48

θ = -10 \pm \sqrt(10)12

This gives two possible values:

θ = -10 + \sqrt(10)12 and θ = -( 10 + \sqrt(10)12)

Step 2: Check Extraneous Roots

When we squared the equation $4 + 5 θ = θ$ $4 + 5\tan\theta = \sec\theta$, we introduced the possibility of extraneous roots where $θ$ $\sec\theta$ might be strictly negative while $4 + 5 θ$ $4 + 5\tan\theta$ is negative, but $α in (-π/2, π/2)$ $\alpha \in (-\pi/2, \pi/2)$ restricts $α gt 0$ $\cos\alpha \gt 0$, hence $α gt 0$ $\sec\alpha \gt 0$. For $α$ $\sec\alpha$ to be positive, $4 + 5 α gt 0$ $4 + 5\tan\alpha \gt 0$. If $α = - 10 + \sqrt(10)12$ (approx $-1.09$ $-1.09$):

4 + 5(-1.09) = 4 - 5.45 = -1.45 lt 0

This contradicts $α gt 0$ $\sec\alpha \gt 0$. Hence, this root is rejected.

Therefore, the only valid solution is:

α = \sqrt(10) - 1012

Pattern Recognition

Whenever you square a trigonometric equation (like converting $\sec$ to $\tan$), always map the proposed roots back to the domain limits to prune out extraneous negative parity roots.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

More Trigonometric Functions Previous-Year Questions — Page 9

Q15 jee_main_2024_31_jan_evening Trigonometric Equations

The number of solutions, of the equation

ex - 2e- x = 2

A. $2$ $2$
B. $more than 2$ $\text{more than } 2$
C. $1$ $1$
D. $0$ $0$

Solution

Core Logic

Let $ex = t$ $e^{\sin x} = t$, where $t > 0$ $t > 0$ because exponential functions are strictly positive. Substitute into the equation:

t - (2)/(t) = 2

t² - 2t - 2 = 0

Solve for $t$ $t$ using the quadratic formula:

t = 2 \pm \sqrt(4 - 4(1)(-2))2 = 1 \pm \sqrt(3)

Since $t > 0$ $t > 0$, we discard $1 - \sqrt(3)$ $1 - \sqrt{3}$. Thus, $t = 1 + \sqrt(3) \approx 2.732$ $t = 1 + \sqrt{3} \approx 2.732$. Now, equate back:

ex = 1 + \sqrt(3) x = ln(1 + \sqrt(3))

We know $e \approx 2.718$ $e \approx 2.718$. Since $1 + \sqrt(3) > e$ $1 + \sqrt{3} > e$, it follows that $ln(1 + \sqrt(3)) > 1$ $\ln(1 + \sqrt{3}) > 1$. But the range of $x$ $\sin x$ is $[-1, 1]$ $[-1, 1]$. Therefore, $x$ $\sin x$ cannot equal a value strictly greater than 1. No real solution exists.

Chapter Mix

Class 11 Maths: Trigonometric Functions Class 12 Maths: Continuity and Differentiability

Q16 jee_main_2024_31_jan_evening Properties of ITFs

a = ⁻¹( (5))

and

b = ⁻¹( (5))

, then

a² + b²

is equal to

A. $4π² + 25$ $4\pi^2 + 25$
B. $8π² - 40π + 50$ $8\pi^2 - 40\pi + 50$
C. $4π² - 20π + 50$ $4\pi^2 - 20\pi + 50$
D. $25$ $25$

Solution

Related Formula

⁻¹( x) = x - 2π for x in [3π/2, 5π/2]

⁻¹( x) = 2π - x for x in [π, 2π]

Core Logic

Evaluate $a = ⁻¹( 5)$ $a = \sin^{-1}(\sin 5)$: The principal branch of $⁻¹ x$ $\sin^{-1} x$ is $[-π/2, π/2]$ $[-\pi/2, \pi/2]$. $5$ $5$ radians is approximately $5 \times 57.3^° \approx 286.5^°$ (in 4th quadrant). The equivalent angle in the principal domain is $5 - 2π$ $5 - 2\pi$. Thus, $a = 5 - 2π$ $a = 5 - 2\pi$.

Evaluate $b = ⁻¹( 5)$ $b = \cos^{-1}(\cos 5)$: The principal branch of $⁻¹ x$ $\cos^{-1} x$ is $[0, π]$ $[0, \pi]$. $5$ $5$ radians is in $[π, 2π]$ $[\pi, 2\pi]$. The equivalent angle is $2π - 5$ $2\pi - 5$. Thus, $b = 2π - 5$ $b = 2\pi - 5$.

Calculate $a² + b²$ $a^2 + b^2$:

a² + b² = (5 - 2π)² + (2π - 5)²

$= 2(5 - 2π)²$ $= 2(5 - 2\pi)^2$

= 2(25 + 4π² - 20π)

= 8π² - 40π + 50

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions

Q15 jee_main_2024_31_jan_morning Properties of Inverse Trigonometric Functions

For

α, β, γ \neq 0

. If

⁻¹α + ⁻¹β + ⁻¹γ = π

and

(α + β + γ)(α - γ + β) = 3 αβ

then

γ

equal to

A. $\sqrt(3)2$ $$\frac{\sqrt{3}}{2}$$
B. $1\sqrt(2)$ $$\frac{1}{\sqrt{2}}$$
C. $\sqrt(3) - 12\sqrt(2)$ $$\frac{\sqrt{3} - 1}{2\sqrt{2}}$$
D. $\sqrt(3)$ $\sqrt{3}$

Solution

Core Logic

Let $⁻¹α = A, ⁻¹β = B, ⁻¹γ = C$ . Given $A + B + C = π$ $A + B + C = \pi$. Since $A = α, B = β, C = γ$ , $α, β, γ$ $\alpha, \beta, \gamma$ act like the side lengths of a triangle divided by $2R$ $2R$ by Sine rule. However, directly dealing with the relation:

(α + β + γ)(α + β - γ) = 3αβ

Step 1: Simplify Algebraic Relation

(α + β)² - γ² = 3αβ

α² + β² + 2αβ - γ² = 3αβ

α² + β² - γ² = αβ

Step 2: Triangle Identification

Divide by $2αβ$ $2\alpha\beta$:

(α² + β² - γ²)/(2αβ) = (1)/(2)

By Cosine Rule, $C = (1)/(2)$ $\cos C = \frac{1}{2}$. Since $C = ⁻¹γ$ $C = \sin^{-1}\gamma$, we know $C = γ$ $\sin C = \gamma$. $C = \sqrt(1 - γ²) = (1)/(2)$ .

Step 3: Final Solution

1 - γ² = (1)/(4) γ² = (3)/(4)

Since $C$ $C$ is an angle of a triangle (or sum equals $π$ $\pi$ and elements are positive limits), $γ = C > 0$ $\gamma = \sin C > 0$.

γ = \sqrt(3)2

Pattern Recognition

The expression $(α + β + γ)(α + β - γ) = 3αβ$ perfectly mirrors the Cosine Rule standard form giving $C = 1/2$ $\cos C = 1/2$.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Trigonometric Functions

If $α$ $\alpha$, $-(π)/(2) lt α lt (π)/(2)$ is the solution of $4 θ+5 θ=1$ $4\cos\theta+5\sin\theta=1$, then the value of $α$ $\tan\alpha$ is

Solution & Explanation

Related Formula

Core Logic

Step 1: Apply Quadratic Formula

Step 2: Check Extraneous Roots

Pattern Recognition

Chapter Mix

Reference Study Guides

More Trigonometric Functions Previous-Year Questions — Page 9

Solution

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Simplify Algebraic Relation

Step 2: Triangle Identification

Step 3: Final Solution

Pattern Recognition

Chapter Mix

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If α$\alpha$, -(π)/(2) lt α lt (π)/(2)$-\frac{\pi}{2} \lt \alpha \lt \frac{\pi}{2}$ is the solution of 4 θ+5 θ=1$4\cos\theta+5\sin\theta=1$, then the value of α$\tan\alpha$ is

Solution & Explanation

Related Formula

Core Logic

Step 1: Apply Quadratic Formula

Step 2: Check Extraneous Roots

Pattern Recognition

Chapter Mix

Reference Study Guides

More Trigonometric Functions Previous-Year Questions — Page 9

Solution

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Simplify Algebraic Relation

Step 2: Triangle Identification

Step 3: Final Solution

Pattern Recognition

Chapter Mix

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If $α$ $\alpha$, $-(π)/(2) lt α lt (π)/(2)$ is the solution of $4 θ+5 θ=1$ $4\cos\theta+5\sin\theta=1$, then the value of $α$ $\tan\alpha$ is