Ncert Solutions For Class 10 Maths

If you are looking NCERT Solutions for Class 10 Maths error free and prepared by experienced mentors. Don’t WORRY!! Here we are providing Free PDF you can get easy with a single click.

SUMMARY 1

tan 𝜃 = cot 𝜃 1

cos 𝜃 = sec 𝜃 1

sin 𝜃 = cosec 𝜃

1

⇒

cot 𝜃 = tan 𝜃

⇒

sec 𝜃 = cos 𝜃

⇒

cosec 𝜃 = sin 𝜃

and

cot 𝜃 = sin 𝜃

1

1

We also observe that

𝑠𝑖𝑛 𝜃

tan 𝜃 = cos 𝜃

cos 𝜃

Trigonometric table: Trigonometric Identities: (i) sin2θ + cos2θ = 1 (ii) 1 + tan2θ = sec2θ (iii) 1 + cot2θ = cosec2θ Trigonometric ratios of complementary angles: Sin (900- θ) = cos θ cos (900- θ) = sin θ Tan (900- θ) = cot θ Cot (900- θ) = tan θ Sec (900- θ) = cosec θ Cosec (900- θ) = sec θ OBJECTIVE 5 sin θ−3 cosθ

1. If 5tan θ = 4, then value of 5 𝑠𝑖𝑛θ+3 cosθ is : 1

a)3

1

𝑏) 6

4

𝑐) 5

2

𝑑) 3

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2. Given 3sin β + 5cos β = 5, then the value of (3cos β- 5sin β)2 is equal to: 9 1 a) 9 𝑏) 5 𝑐) 3 3. If tan θ= 4, then (𝑠𝑖𝑛3𝜃 𝑐𝑜𝑠𝜃

tan θ +𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃

) is equal to:

𝑏)2√2

a) 0

1

𝑑) 9

𝑐)√2

𝑑)1

𝜋

4. As x increases from 0 to 2 , the value of cos x : a) Increases b) decreases c) remains constant

d) increases, then decreases

5. Value of x from the equation 𝜋

𝜋

xsin 6 cos2 4 =

𝜋 6

𝜋 3

𝜋 4 𝜋 𝜋 𝑐𝑜𝑠𝑒𝑐 2 𝑐𝑜𝑠𝑒𝑐 4 6

𝑐𝑜𝑡 2 𝑠𝑒𝑐 𝑡𝑎𝑛

a) 4

is :

b) 6

c) -2

d) 0

6. The area of a triangle is 12 sq. cm. Two sides are 6 cm and 12 cm. The included angle is : 1 1 1 1 a) cos −1 (3) b) cos −1 (6) c) sin−1 (6) d) sin−1 (3) 7. If α + β = 900 and α = 2β, then cos2𝛼 + sin2β equals to: 1 a) 2 b) 0 c) 1

d) 2

SUBJECTIVE 1.

If A + B = 900, prove that: 𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵+𝑡𝑎𝑛𝐴𝑐𝑜𝑡𝐵

√

𝑠𝑖𝑛𝐴𝑠𝑒𝑐𝐵

𝑠𝑖𝑛2 𝐵

− 𝑐𝑜𝑠2 𝐴 = tan 𝐴

Prove the following (Q 2 to Q5) 2.

3.

4.

5.

𝑡𝑎𝑛𝜃 1−𝑐𝑜𝑡𝜃

𝑐𝑜𝑡𝜃

+ 1−𝑡𝑎𝑛𝜃 = sec𝜃𝑐𝑜𝑠𝑒𝑐𝜃 + 1

𝑠𝑖𝑛𝐴+𝑐𝑜𝑠𝐴 𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴

𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴

+ 𝑠𝑖𝑛𝐴+𝑐𝑜𝑠𝐴 =

2

2

𝑠𝑖𝑛2 𝐴−𝑐𝑜𝑠2 𝐴

= 1−2𝑐𝑜𝑠2𝐴

(𝑠𝑖𝑛8 𝜃 − 𝑐𝑜𝑠 8 𝜃) = (𝑠𝑖𝑛2 𝜃 − 𝑐𝑜𝑠 2 𝜃)( 1 − 𝑠𝑖𝑛2 𝜃𝑐𝑜𝑠 2 𝜃) 𝑡𝑎𝑛𝜃+𝑠𝑒𝑐𝜃−1

= 𝑡𝑎𝑛𝜃−𝑠𝑒𝑐𝜃+1

1+𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

6.

If x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, then prove that: x2+y2+z2=r2.

7.

Prove that:

8.

Evaluate: tan 70 tan 230 tan 600 tan 670 tan 830 + tan 360 + sin 200 sec 700- 2

1 𝑠𝑒𝑐𝑥−𝑡𝑎𝑛𝑥

1

1

1

− 𝑐𝑜𝑠𝑥 = 𝑐𝑜𝑠𝑥 − 𝑠𝑒𝑐𝑥+𝑡𝑎𝑛𝑥 cot 540

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2

2

5

9.

Evaluate: 3cosec2580 - 3cot 580tan 320 - 3tan 130 tan 370 tan 450 tan 530 tan 770.

10.

If 5x = sec θ and 𝑥 = tanθ, find the value of 5(𝑥 2 − 𝑥 2 ).

11.

Find the value of sec 600 geometrically.

12.

Prove the following: sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A.

5

1

Answers: Objective: 1) b

2) a

3) d

Subjective: 8)√3

9)-1

10) 5

4) b 1

5) b

6) d

7) a

11) 2

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