If x + 2 x = 1, x (0, (pi/2)), then (12 x + 12 x) + 3 (10 x + 10 x + 8 x + 8 x) + (6 x + 6 x) is equal to

Year	2026	2025	2024	Total
Questions	10	10	7	27

Q2 jee_main_2026_23_january_evening Trigonometric Identities

Let

(π)/(2)<θ<π

and

θ=- 12\sqrt(2)

. Then the value of

((15θ)/(2))( 8θ+ 8θ)+ ((15θ)/(2))( 8θ- 8θ)

is equal to:

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

Solution

Related Formula

Compound angle formulas:

(A - B) = A B - A B

(A - B) = A B + A B

Core Logic

Trigonometric Identities diagram for Q2 - JEE Main 2026 Evening

Expanding the given expression:

= ((15θ)/(2)) 8θ + ((15θ)/(2)) 8θ + ((15θ)/(2)) 8θ - ((15θ)/(2)) 8θ

Rearranging terms to form standard identities:

=[ ((15θ)/(2)) 8θ - ((15θ)/(2)) 8θ] + [ ((15θ)/(2)) 8θ + ((15θ)/(2)) 8θ]

= ((15θ)/(2) - 8θ) + ((15θ)/(2) - 8θ)

= (-(θ)/(2)) + (-(θ)/(2))

= (θ)/(2) - (θ)/(2)

Step 1: Finding Trigonometric Values

Given $θ = - 12\sqrt(2)$ $\cot\theta = -\frac{1}{2\sqrt{2}}$ and $θ in ((π)/(2), π)$ $\theta \in (\frac{\pi}{2}, \pi)$. This implies $θ = 2\sqrt(2)3$ $\sin\theta = \frac{2\sqrt{2}}{3}$.

We need to evaluate $(θ)/(2) - (θ)/(2)$ . We know:

( (θ)/(2) - (θ)/(2))² = 1 - θ

Since $(π)/(2) < θ < π$ $\frac{\pi}{2} < \theta < \pi$, we have $(π)/(4) < (θ)/(2) < (π)/(2)$ . In this quadrant, $(θ)/(2) > (θ)/(2)$ . Therefore, $(θ)/(2) - (θ)/(2) = -\sqrt(1 - θ)$ .

Step 2: Final Calculation

- 1 - 2\sqrt(2)3 = - 3 - 2\sqrt(2)3

Notice that $3 - 2\sqrt(2) = (\sqrt(2) - 1)²$ $3 - 2\sqrt{2} = (\sqrt{2} - 1)^2$.

=- \sqrt(2) - 1\sqrt(3) = 1 - \sqrt(2)\sqrt(3)

Pattern Recognition

Recognize the expanded forms of $(A-B)$ $\sin(A-B)$ and $(A-B)$ $\cos(A-B)$ hiding within the products. Also, carefully determine the sign of $(θ/2) - (θ/2)$ $\cos(\theta/2) - \sin(\theta/2)$ based on the half-angle quadrant.

Chapter Mix

Class 11 Maths: Trigonometric Functions

Q13 jee_main_2026_23_january_evening Maximum and Minimum Values

The least value of

( ²θ - 6 θ θ + 3 ²θ + 2)

is

A. $-1$ $-1$
B. $4+\sqrt(10)$ $4+\sqrt{10}$
C. $4-\sqrt(10)$ $4-\sqrt{10}$
D. $1$ $1$

Solution

Related Formula

The expression $A x + B x + C$ $A\sin x + B\cos x + C$ lies in the range:

[C - \sqrt(A² + B²), C + \sqrt(A² + B²)]

Core Logic

Let $f(θ) = ²θ - 6 θ θ + 3 ²θ + 2$ . Convert all quadratic terms to multiple angles:

f(θ) = (1 + 2θ)/(2) - 3(2 θ θ) + 3((1 - 2θ)/(2)) + 2

f(θ) = (1)/(2) + (1)/(2) 2θ - 3 2θ + (3)/(2) - (3)/(2) 2θ + 2

Step 1: Simplifying Expression

f(θ) = ((1)/(2) + (3)/(2) + 2) - 3 2θ + ((1)/(2) - (3)/(2)) 2θ

f(θ) = 4 - 3 2θ - 2θ

Step 2: Finding Extremes

The expression is of the form $C + A 2θ + B 2θ$ $C + A\sin 2\theta + B\cos 2\theta$, where $C=4, A=-3, B=-1$ $C=4, A=-3, B=-1$. The minimum value is $C - \sqrt(A² + B²)$ $C - \sqrt{A^2 + B^2}$:

4 - \sqrt((-3)² + (-1)²) = 4 - \sqrt(9 + 1) = 4 - \sqrt(10)

The maximum value is $C + \sqrt(A² + B²) = 4 + \sqrt(10)$ $C + \sqrt{A^2 + B^2} = 4 + \sqrt{10}$. Therefore, the least value is $4 - \sqrt(10)$ $4 - \sqrt{10}$.

Pattern Recognition

Every quadratic trigonometric expression in $θ$ $\sin\theta$ and $θ$ $\cos\theta$ must be linearly transformed into $(2θ)$ $\sin(2\theta)$ and $(2θ)$ $\cos(2\theta)$ terms to utilize the standard bound formula.

Chapter Mix

Class 11 Maths: Trigonometric Functions

Q8 jee_main_2026_24_january_morning Trigonometric Identities and Simplification

The value of

\sqrt(3)cosec20°- 20° 20° 40° 60° 80°

is equal to

A. $32$ $32$
B. $16$ $16$
C. $64$ $64$
D. $12$ $12$

Solution

Related Formula

θ (60^° - θ) (60^° + θ) = (1)/(4) 3θ

A B - A B = (A - B)

Core Logic

Evaluate the denominator first: $D = 20^° 40^° 60^° 80^°$ . Using $θ (60^°-θ) (60^°+θ)$ with $θ = 20^°$ $\theta = 20^\circ$:

D = 60^° [ (1)/(4) (3 \times 20^°) ] = (1)/(2) \cdot (1)/(4) \cdot (1)/(2) = (1)/(16)

Step 1: Simplify Numerator

N = \sqrt(3) 20^° - (1)/( 20^°)

N = \sqrt(3) 20^° - 20^° 20^° 20^°

Multiply and divide by 2:

N = 2 ( \sqrt(3)2 20^° - (1)/(2) 20^° )(1)/(2) 40^°

N = (4 ( 60^° 20^° - 60^° 20^°))/( 40^°)

N = (4 (60^° - 20^°))/( 40^°) = (4 40^°)/( 40^°) = 4

Step 2: Combine Terms

E = (N)/(D) = (4)/(1/16) = 64

Pattern Recognition

The expression $\sqrt(3)cosecθ - θ$ simplifies instantly to $4$ $4$ by creating a sine difference angle identity. Denominators of $20^° 40^° 80^°$ $\cos20^\circ\cos40^\circ\cos80^\circ$ universally yield $(1)/(8)$ $\frac{1}{8}$.

Chapter Mix

Class 11 Maths: Trigonometric Functions

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

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

If x + ² x = 1$\sin x + \sin^2 x = 1$, x in (0, (π)/(2))$x \in \left(0, \frac{\pi}{2}\right)$, then ( ¹² x + ¹² x) + 3 ( ¹⁰ x + ¹⁰ x + ⁸ x + ⁸ x) + ( ⁶ x + ⁶ x)$(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x)$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify the Expression

Step 2: Apply the Cubic Identity

Pattern Recognition

Chapter Mix

Reference Study Guides

More Trigonometric Functions Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Finding Trigonometric Values

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplifying Expression

Step 2: Finding Extremes

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Simplify Numerator

Step 2: Combine Terms

Pattern Recognition

Chapter Mix

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: Algebraic Simplification

Step 2: Conclusion

Step 3: Final Answer

Chapter Mix

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If $x + ² x = 1$ $\sin x + \sin^2 x = 1$, $x in (0, (π)/(2))$ $x \in \left(0, \frac{\pi}{2}\right)$, then $( ¹² x + ¹² x) + 3 ( ¹⁰ x + ¹⁰ x + ⁸ x + ⁸ x) + ( ⁶ x + ⁶ x)$ is equal to