NCERT Solutions
Class 10 Maths
Introduction to Trigonometry
Electric charges

Ex.8.1 Q.1

In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:

(1) sin A, cos A                                   

(2) sin C, cos C

 

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Ex.8.1 Q.2

In Fig. 8.13, find tan P – cot R.

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Ex.8.1 Q.3

If sin A = , calculate cos A and tan A.

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Ex.8.1 Q.4

Given 15 cot A = 8, find sin A and sec A.

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Ex.8.1 Q.5

Given sec θ = , calculate all other trigonometric ratios.

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Ex.8.1 Q.6

If A and B are acute angles such that cos A = cos B, then show that A = B.

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Ex.8.1 Q.7

If cot θ =   evaluate:

(1) {(1 + sin θ) (1 - sin θ)} ÷ {(1 + cos θ) (1 - cos θ)}                         

(2) cot2 θ

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Ex.8.1 Q.8

If 3 cot A = 4, check whether (1 - tan2 A) ÷ (1 + tan2 A) = cos2 A – sin2 A or not.

 

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Ex.8.1 Q.9

In triangle ABC, right-angled at B, if tan A = , find the value of:

(1) sin A cos C + cos A sin C                                 

(2) cos A cos C – sin A sin C        

 

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Ex.8.1 Q.10

In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm.

Determine the values of sin P, cos P and tan P.

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Ex.8.1 Q.11

State whether the following are true or false. Justify your answer.

(1) The value of tan A is always less than 1.

(2) sec A =  for some value of angle A.

(3) cos A is the abbreviation used for the cosecant of angle A.

(4) cot A is the product of cot and A.

(5) sin θ =  for some angle θ.

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Ex.8.2 Q.1

Evaluate the following:

(1) sin 600 cos 300 + sin 300 cos 600                      

(2) 2 tan2 450 + cos2 300 – sin2 600

(3) cos 450 ÷ (sec 300 + cosec 300)   

(4) (sin 300 + tan 450 – cosec 600) ÷ (sec 300 + cos 600 + cot 450)

(5) (5 cos2 600 + 4 sec2 300 - tan2 450) ÷ (sin2 300 + cos2 300)

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Ex.8.2 Q.2

Choose the correct option and justify your choice:

(1) 2 tan 300 ÷ (1 + tan2 300) =

(A) sin 600                    

(B) cos 600                          

(C) tan 600                       

(D) sin 300

(2) (1 – tan2 450) ÷ (1 + tan2 450) =

(A) tan 900                            

(B) 1                             

(C) sin 450                              

(D) 0

(3) sin 2A = 2 sin A is true when A =

(A) 00                                     

(B) 300                         

(C) 450                                     

(D) 600

(4) 2 tan 300 ÷ (1 – tan2 300) =

(A) cos 600                            

(B) sin 600                   

(C) tan 600                              

(D) sin 300

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Ex.8.2 Q.3

If tan (A + B) = √3 and tan (A – B) = ; 00 < A + B ≤ 900; A > B, find A and B.

 

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Ex.8.2 Q.4

State whether the following are true or false. Justify your answer.

(1) sin (A + B) = sin A + sin B.

(2) The value of sin θ increases as θ increases.

(3) The value of cos θ increases as θ increases.

(4) sin θ = cos θ for all values of θ.

(5) cot A is not defined for A = 00.

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Ex.8.3 Q.1

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

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Ex.8.3 Q.2

Write all the other trigonometric ratios of A in terms of sec A.

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Ex.8.3 Q.3

Choose the correct option. Justify your choice.

(1) 9 sec2 A – 9 tan2 A =

(A) 1                     

(B) 9                   

(C) 8                                 

(D) 0

(2) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =

(A) 0                     

(B) 1                   

(C) 2                                 

(D) -1

(3) (sec A + tan A) (1 – sin A) =

(A) sec A              

(B) sin A             

(C) cosec A                     

(D) cos A

(4) (1 + tan2 A) ÷ (1 + cot2 A) =

(A) sec2 A            

(B) –1                 

(C) cot2 A                       

(D) tan2 A

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Ex.8.3 Q.4

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(1) (cosec θ – cot θ)2        

(2)  +  = 2 sec A

(3)   +   = 1 + sec θ + cosec θ

[Hint: Write the expression in terms of sin θ and cos θ]

(4)   = sin2 A ÷ (1 – cos A)

[Hint: Simplify LHS and RHS separately]

(5) (cos A – sin A + 1) ÷ (cos A + sin A - 1) = cosec A, using the identity cosec2 A = 1 + cot2 A.

(6) √ {(1 + sin A) ÷ (1 – sin A)} = sec A + tan A

(7) (sin θ – 2 sin3 θ ÷ (2cos3 θ – cos θ) = tan θ

(8) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

(9) (cosec A – sin A) (sec A – cos A) = 

[Hint: Simplify LHS and RHS separately]

(10) (1 + tan2 A) ÷ (1 + cot2 A) = (1 - tan2 A) ÷ (1 - cot2 A) = tan2 A

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Complete NCERT Solutions: Classes 6 to 12, All Chapters

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