Class 10

Prove that sinA(1+tanA)+cosA(1+cotA) . . .

Question : Prove that sinA(1+tanA)+cosA(1+cotA) = secA+cosecA

Doubt by Afifa

Solution : OR

sinA(1+tanA)+cosA(1+cotA) = secA+cosecA

LHS
sinA(1+tanA)+cosA(1+cotA)
=sinA(1+sinA/cosA) + cosA(1+cosA/sinA)

[∵tanA=sinA/cosA, cotA=cosA/sinA]
=sinA[(cosA+sinA)]/cosA+cosA[(sinA+cosA)]/sinA
=sinA[(sinA+cosA)]/cosA+cosA[(sinA+cosA)]/sinA

=sinA+cosA[sinA/cosA+cosA/sinA]
=sinA+cosA[(sin²A+cos²A)/sinAcosA]
=sinA+cosA[1/sinAcosA]
[∵sin²A+cos²A=1]

=(sinA+cosA)/sinAcosA
=sinA/sinAcosA + cosA/sinAcosA
= 1/cosA + 1/sinA

= secA+cosecA
[∵1/cosA=secA and 1/sinA=cosecA]
= RHS

LHS=RHS
Hence Proved.

In ΔABC, ∠ACB=90° and DCEF is a square . . .

Question : In ΔABC, ∠ACB=90° and DCEF is a square, as shown in figure. Prove that DE²=AD×BF

Doubt by Afifa

Solution :

DCEF is a square.
DE=EF=FC=CD (All sides of square are equal)

In ΔACB and ΔADE
∠A=∠A (Common)
∠ACB=∠ADE (Each 90°)
ΔACB~ΔADE (By AA Similarity Criteria) — (1)

In ΔACB and ΔEFB
∠B=∠B (Common)
∠ACB=∠EFB (Each 90°)
ΔACB~ΔEFB (By AA Similarity Criteria) — (2)

From equation (1) & (2)

ΔADE~ΔEFB

AD/EF=ED/BF (By CPCT)
AD×BF=EF×ED
AD×BF=EF×DE
AD×BF=DE×DE [∵EF=DE]
AD×BF=DE²
DE²=AD×BF

Hence Proved.

If the ratio of the sum of the first n terms of two A.P.s is (7n+1) . . .

Question : If the ratio of the sum of the first n terms of two A.P.s is (7n+1):(4n+27), then find the ratio of their 9th terms.

Doubt by Shaurya

Solution :

Let us consider two AP's having first term and common difference as a and a' and d and d' respectively.

S_n:S_n'=(7n+1):(4n+27)
n/2[2a+(n-1)d] : n/2[2a'+(n-1)d']=(7n+1):(4n+27)
[2a+(n-1)d]:[2a'+(n-1)d']=(7n+1):(4n+27)

2[a+{(n-1)/2}d]:2[a'+{(n-1)/2}d']=(7n+1):(4n+27)
[a+{(n-1)/2}d]:[a'+{(n-1)/2}d']=(7n+1):(4n+27) — (1)

Now in order to find the ratio of 9th term
{(n-1)/2} should be equal to 8
{(n-1)/2}=8
n-1=16
n=16+1
n=17

Putting n=17 in equation (1)

a+8d:a'+8d=[7(17)+1]:[4(17)+27]
a+(9-1)d : a'+(9-1)d'=[119+1] : [68+27]
a₉:a'₉=120:95
a₉:a'₉=24:19

Hence, the ratio of 9th term is 24:19

Divide 32 into four parts which are in A.P. such that the product

Question : Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7:15.

Doubt by Shaurya

Solution :

Let the four parts which are in AP are
a₁=a-3d
a₂=a-d
a₃=a+d
a₄=a+3d

ATQ

a₁+a₂+a₃+a₄=32

a-3d+a-d+a+d+a+3d=32

4a=32

a=32/4

a=8 — (1)

(a₁×a₄):(a₂×a₃) = 7:15

(a₁×a₄)/(a₂×a₃) = 7/15

(a-3d)(a+3d)/(a-d)(a+d)=7/15
(a²-9d²)/(a²-d²)=7/15
(a²-9d²)/(a²-d²)=7/15
15(a²-9d²)/=7×(a²-d²)
15a²-135d²=7a²-7d²
15a²-7a²=135d²-7d²
8a²=128d²
8(8)²/128=d² [From equation (1)]
8×64/128=d²
8/2=d²
4=d²
d²=4
d=±√4

d=±2

When d=+2, a=8
a₁=a-3d = 8-3(2) = 8-6 = 2

a₂=a-d = 8-2 = 6
a₃=a+d = 8+2 = 10
a₄=a+3d = 8+3(2) = 8+6 = 14

When d=-2, a=8

a₁=a-3d = 8-3(-2) = 8+6 = 14

a₂=a-d = 8-(-2) = 8+2 = 10
a₃=a+d = 8+(-2) = 8-2 = 6
a₄=a+3d = 8+3(-2) = 8-6 = 2

Hence, the required four parts are 2, 6, 10, 14 or 14, 10, 6, 2

If ΔABC~ΔPQR in which AB=6cm . . .

Question : If ΔABC~ΔPQR in which AB=6cm, BC=4cm, AC=8cm and PR=6 cm, then find the length of (PQ+QR).

Doubt by Shaurya and Prateek

Solution :

AB=6cm (Given) BC=4cm (Given) AC=8cm (Given) PR=6 cm (Given)

ΔABC~ΔPQR (Given) AB/PQ=BC/QR=AC/PR (Corresponding sides of similar triangle are proportional) 6/PQ=4/QR=8/6 6/PQ=4/QR=4/3 Now, 6/PQ=4/3 4PQ=18 PQ=18/4 PQ=4.5 cm

Also, 4/QR=4/3 QR=3 cm PQ+QR =4.5+3 =7.5 cm

∴PQ+QR=7.5 cm

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.

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