Assertion (A): If cosA+cos2A=1, then sin2A+sin4A . . .

Question : Assertion (A): If cosA+cos²A=1, then sin²A+sin⁴A=1

Reason (R): sin²A+cos²A=1

Choose the correct option:

(A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

(B)Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

(C)Assertion (A) is true but reason (R) is false.

(D)Assertion (A) is false but reason (R) is true.

CBSE Sample Question Paper 2026

Solution : 

cosA+cos²A=1 (Given) —(1) 
cosA=1-cos²A
cosA=sin²A — (2) 
[ ∵ sin²θ+cos²θ=1]

Now 
sin²A+sin⁴A

=sin²A+(sin²A)²
=cosA+(cosA)²
=cosA+cos²A

=1 [From equation (1)]

Hence, sin²A+sin⁴A=1

We have proved this by using the identity sin²A+cos²A=1, which is given in assertion.

So, (A)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A), will be the correct option.

If 2sin5x=√3, 0°<=x<=90°, then x is equal to . . .

Question : If 2sin5x=√3, 0°≤x≤90°, then x is equal to 

(A) 10°
(B) 12°
(C) 20°
(D) 50°

CBSE Sample Question Paper 2026

Solution : 

2sin5x=√3
sin5x=√3/2
sin5x=sin60°
5x=60°
x=60°/5
x=12°

Hence, (B) 12°, would be the correct option. 

If secθ+tanθ=x, then secθ-tanθ will be . . .

Question : If secθ+tanθ=x, then secθ-tanθ will be

(A) x
(B) x²
(C) 2/x
(D) 1/x

CBSE Sample Question Paper 2026

Solution : 

secθ+tanθ=x
Reciprocating both side

1/(secθ+tanθ) = 1/x
sec²θ-tan²θ/(secθ+tanθ)=1/x [∵1+tan²θ=sec²θ]
[(secθ+tanθ)(secθ-tanθ)]/(secθ+tanθ)=1/x
[∵a²-b²=(a+b)(a-b)]
secθ-tanθ=1/x

Hence, (D) 1/x, would be the correct option. 

If sinθ+cosθ=√3, then prove that . . .

Question : If sinθ+cosθ=√3, then prove that tanθ+cotθ=1

CBSE Sample Question Paper 2026

Solution : 

sinθ+cosθ=√3
S.B.S
(sinθ+cosθ)²=(√3)²
sin²θ+cos²θ+2sinθcosθ=3
1+2sinθcosθ=3 [∵sin²θ+cos²θ=1]
2sinθcosθ=3-1
2sinθcosθ=2
sinθcosθ=2/2
sinθcosθ=1 — (1) 

To Prove : 
tanθ+cotθ=1

Proof : 
LHS:

tanθ+cotθ
=sinθ/cosθ + cosθ/sinθ
=[sin²θ+cos²θ]/[sinθcosθ]
=1/[sinθcosθ] [∵sin²θ+cos²θ=1]
= 1/1 [From Equation (1)]
=1
=RHS

LHS=RHS
Hence Proved.

[CBSE 2025] The students of a class are made to stand equally in rows. If 3 students are extra in . . .

Question : The students of a class are made to stand equally in rows. If 3 students are extra in each row, there would be 1 row less. If 3 students are less in a row, there would be 2 more rows. Find the number of students in the class.

CBSE 2025

Solution :
Let the number of students in each row = x
and the number of rows = y

Total number of students = xy

ATQ

(x+3)(y-1)=xy
x(y-1)+3(y-1)=xy
xy-x+3y-3=xy
-x+3y-3=0
-x+3y=3 — (1)

Also

(x-3)(y+2)=xy
x(y+2)-3(y+2)=xy
xy+2x-3y-6=xy
2x-3y-6=0
2x-3y=6 — (2)

Solving equation (1) and (2)

-x+3y=3
2x-3y=6
--------------
x+0y=9
--------------

x=9

putting in equation (1)
-9+3y=3
3y=3+9
3y=12
y=12/3
y=4

Number of students in each row = 9
and the number of rows = 4

Total number of students
=xy
=9×4
=36

Hence, the total number of students in the class is 36.

[CBSE 2025] The perimeter of a rectangle is 70 cm. The length of the rectangle is 5 cm . . .

Question : The perimeter of a rectangle is 70 cm. The length of the rectangle is 5 cm more than twice its breadth. Express the given situation as a system of linear equations in two variables and hence solve it. 

CBSE 2025

Solution : 

Let the length and breath of a rectangle be x cm and y cm respectively. 

ATQ

Perimeter of rectangle = 2(l+b)
70 = 2(x+y)
70/2=x+y
35=x+y
x+y=35 — (1) 

Also, 

x=5+2y
x-2y=5 — (2) 

Solving equation (1) and (2) 

x+y=35

x-2y=5
- +     -
---------------
0 +3y=30
---------------

3y=30

y=30/3
y=10

putting in equation (1)

x+10=35
x=35-10
x=25

Hence, length = 25 cm and breadth = 10 cm 

The two angles of a right angled triangle other than . . .

Question : The two angles of a right angled triangle other than 90° are in the ratio 2:3. Express the given situation algebraically as a system of linear equations in two variables and hence solve it. 

CBSE 2025

Solution : 

Let the measures of the two angles be x and y. 

ATQ
x+y+90°=180° (ASP)
x+y=180°-90°
x+y=90° — (1) 

Also 
x:y=2:3
x/y=2/3
3x=2y
3x-2y=0 — (2) 

Solving equation (1) and (2) 

[x+y=90°]×2

[3x-2y=0]×1

2x+2y=180°
3x-2y=0
----------------
5x+0y=180
----------------

5x=180
x=180/5
x=36°

putting in equation (1)

36°+y=90°
y=90°-36°
y=54°

Hence, the required values of two angles is 36° and 54°.

The area of a rectangle gets reduced by 9 square units . . .

Question : The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Doubt by Noor

Solution : 

Let length of the rectangle be x units 

and breadth of the rectangle be y units

Area of rectangle = l×b
= xy square units. 

ATQ

(x-5)×(y+3)=(xy-9)
x(y+3)-5(y+3)=xy-9
xy+3x-5y-15=xy-9

3x-5y-15=-9

3x-5y=-9+15
3x-5y=6 — (1) 

Also

(x+3)×(y+2)=(xy+67)
x(y+2)+3(y+2)=xy+67
xy+2x+3y+6=xy+67
2x+3y+6=67
2x+3y=67-6
2x+3y=61 — (2) 

Solving equations (1) and (2) 

3x-5y=6

2x+3y=61

[3x-5y=6 ]×2

[2x+3y=61]×3

6x-10y=12

6x+9y=183
-   -       - 
-----------------
0x-19y=-171
-----------------

-19y=-128
19y=128
y=128/19
y=9

putting y=9 in equation (1) 
3x-5(9)=6
3x-45=6
3x=6+45
3x=51
x=51/3
x=17

Hence, length of rectangle is 17 units and breadth of rectangle is 9 units. 

[CBSE 2025] √2x+√3y = 5, √3x-√8y =-√6 Solve for x and y . . .

Question : Solve for x and y :

√2x+√3y = 5

√3x-√8y =-√6

CBSE 2025

Solution : 

√2x+√3y = 5 — (1)

√3x-√8y =-√6 — (2) 

[√2x+√3y = 5]×√3

[√3x-√8y =-√6]×√2

√6x+3y=5√3

√6x-4y=-√12
-    +      +
---------------------------
0x  + 7y =5√3+√12
---------------------------

7y=5√3+2√3
7y=7√3
y=7√3/7
y=√3

Putting in equation (1) 

√2x+√3(√3) = 5
√2x+3=5
√2x=5-3
√2x=2
x=2/√2
x=√2

Hence, x=√2 and y=√3

[CBSE 2025] 30x+44y=10, 40x+55y=13 Solve the following system of equations algebraically . . .

Question : Solve the following system of equations algebraically : 

30x+44y=10 

40x+55y=13

CBSE 2025

Solution : 

30x+44y=10 — (1)

40x+55y=13 — (2) 

[30x+44y=10]×4

[40x+55y=13]×3

120x+176y=40
120x+165y=39

-       -          - 
----------------------
0x   +  11y = 1
----------------------

11y=1
y=1/11

Putting in equation (1) 
30x+44y=10
30x+44(1/11)=10

30x+4=10
30x=10-4
30x=6
x=6/30

x=1/5
Hence, x=1/5 and y=1/11.