NCERT Class 10 Math Chapter 8 Important Questions Answer – Introduction to Trigonometry

 Class 10 Chapter Introduction to Trigonometry Subject Math Category Important Question Answer

Class 10 Math Chapter 8 Important Question Answer

Q1. Prove that :

Ans. LHS

RHS

As we know sin2A + cos2A = 1. Therefore, 1 – cos2A = sin2A

LHS = RHS

Hence proved.

Q2. Prove that :   = 1 + secθ cosecθ  Most Important

Ans.  LHS

=

Using a3 – b3 = (a – b)(a2 + b2 + ab)

=

RHS

1 + secθ cosecθ =

LHS = RHS

Hence proved.

Q3. Prove that :  (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A

Ans. LHS

(sin A + cosec A)2 + (cos A + sec A)2 = sin2A + cosec2A + 2sinAcosecA + cos2A + sec2A + 2cosAsecA

[As we know sin2A + cos2A = 1, 1 + tan2A = sec2A, 1 + cot2A = cosec2A, ]

= sin2A + cos2A +  cosec2A + 2 + + sec2A + 2

= 1 + 1 + cot2A + 1 + tan2A + 4

= 7 + tan2A + cot2A

LHS = RHS

Hence proved.

Q4. Prove that  , using the identity cosec2A- cot2A = 1.

Ans. Since we will apply the identity involving cosecA and cotA, let us first convert the LHS in terms of coseA and cotA by dividing numerator and denominator by sinA.

LHS

=

=

=

LHS = RHS

Hence proved.

Q5. Prove that:      Most Important

Ans. Since, in RHS angles are given in secθ and tanθ, we will convert the LHS in terms of secθ and tanθ by dividing numerator and denominator of LHS by cosA.

LHS

=

=

LHS = RHS

Hence proved.

Q6. Prove that:

Ans. Since we will apply the identity involving cosecA and cotA, let us first convert the LHS in terms of coseA and cotA by dividing numerator and denominator by sinA.

LHS

=

=

=

= = cosecA + cotA

LHS = RHS

Hence proved.

Q7. Prove that :

Ans. LHS

=

LHS = RHS

Hence proved.

Q8. Prove that :

Ans. LHS

=

LHS = RHS

Hence proved.

Q9. Prove that:

Ans. LHS

=

RHS

LHS = RHS

Hence proved.

Q10. Prove that:

Ans. LHS

=

=

LHS = RHS

Hence proved.

Q11. Prove that:  (cosecA – sinA) (secA – cosA) (tanA + cotA) = 1

Ans. LHS

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

=

=

=

=  1

LHS = RHS

Hence proved.

Q12. Prove that :

Ans. Since, in RHS angles are given in secθ and tanθ, we will convert the LHS in terms of secθ and tanθ by dividing numerator and denominator of LHS by cosA.

LHS

=

=

=

LHS = RHS

Hence proved.

Q13. Find the value of sin 45° + cos 45°. Most Important

Ans. sin 45° + cos 45°=

Q14. If sin A = , find the value of cos A.

Ans. SinA = =

Using pythagoras theorem, H2  = P2 + B2

52 = 42 + B2

B2 = 9

B = 3

CosA = =

Q15. If tan (A + B) =  and tan(A – B) = , 0° < A + B ≤ 90°, A > B, then find the value of A and B. Most Important

Ans. As we know tan60° =

Therefore, A + B = 60°  —– (i)

Similarly, tan30° =

Therefore, A – B = 30°  —— (ii)

Adding eq (i) and (ii) we get

2A = 90°

A = 45°

Putting value of A in eq (i) we get,

B = 15°

Q16. If sin(A + B) = and sin(A – B) = , 0° < A + B ≤ 90°, A > B, then find A and B.

Ans. As we know sin60° =

Therefore, A + B = 60°  —– (i)

Similarly, sin30° =

Therefore, A – B = 30°  —— (ii)

Adding eq (i) and (ii) we get

2A = 90°

A = 45°

Putting value of A in eq (i) we get,

B = 15°

Q17. The value of 1 + tan2 θ = ___________.

Ans. sec2θ

Q18. Find the value of

Ans. = =

Q19. Prove that :

Ans. LHS

= = RHS

LHS = RHS

Hence proved.

Q20. Prove that : sec A (1 – sin A) (sec A + tan A) = 1

Ans. LHS

sec A (1 – sin A) (sec A + tan A) =

=

=

LHS = RHS

Hence proved.

Q21. In ΔOPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm. Determine the values of sin Q and cos Q.

Ans.

Given, OP = 7cm

Let PQ = x cm, then OQ = (x + 1) cm

In right ΔOPQ,

Using pythgoras theorem,

OP2 + PQ2 = OQ2

72 + x2 = (x+1)2

49 + x2 = x2 + 1 + 2x

48 = 2x

x = 24 cm

PQ = 24 cm, and OQ = 24 + 1 = 25 cm

Now, SinQ =

CosQ =

Q22. If sin(A-B) = , cos(A+B) = , 0° < A + B ≤ 90°, A > B, find A and B.

Ans. As we know

sin30o =

Therefore, A – B = 30  —- (i)

Similarly, cos60o =

Therefore, A + B = 60 o  —– (ii)

Adding eq (i) and (ii) we get

2A = 90 o

A = 45 o

Putting the value of A in eq (ii) we get

B = 15 o

Q23. In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P.

Ans.

Given, PQ = 5 cm, ΔPQR is a right angled triangle.

Let PR = x cm,

Then QR = (25 – x) cm

In right ΔPQR,

Using pythagoras theorem,

PQ2 + QR2 = PR2

52 + (25 – x)2 = x2

25 + 625 – 50x + x2 = x2

650 = 50x

x = 13 cm

Now, x = PR = 13cm and QR = 25 – x = 25 – 13 = 12 cm

SinP =

Q24. “cosθ =  for some angle θ” (true / false)

Ans. As base of triangle can never be greater than or equal to hypotenuse.

Therefore, given value of cosθ is false.

False.

Q25. “sinθ= cosθ for all values of θ” (true / false)

Ans. False.

As sin0o ≠ cos0o

Q26. “The value of tanA is always lies in between -1 and 1” (true / false)

Ans. As tanA = and when cosA = 0

Then, tanA will be not defined.

Hence given statement is false.

Q27. “cotA is not defined for A = 0° ” (true / false)

Ans. True.

As cotA = = not defined.

Q28. “The value of secA is always lies between -1 and 1” (true / false)

Ans. False.

At angle 60o value of sec is 2 which doesn’t lie in this range.

Q29. “cosec A is the product of cosec and A” (true / false)

Ans. False.

Q30. “The value of cosecθ lies  ≥ 1 and ≤ 1 ” (true / false)

Ans. True.

Q31. “sinθ = for some angle θ ” (true / false)

Ans. False.

Because value of perpendicular can never be greater than or equal to hypotenuse.

Q32. Evaluate:

Ans.

[Using, sin2θ + cos2θ = 1]

Q33. Evaluate:

Ans.

=

Q34. If sec θ = , then find sin θ.

Ans. secθ = ,

Using pythagoras theorem, H2 = P2 + B2, where H is hypoteneus, P is perpendicular and B is base.

132 = P2 + 122

P2 = 169 – 144 = 25

P = 5

Now, sinθ =

Q35. The value of 2tan2 45° + cos2 30° – sin2 60° is

Ans.

=

= 2

Q36. sin 60° cos 30° + sin 30° cos 60° is equal to _________.

Ans.

=

Q37. The value of cos2 θ + sin2 θ = __________.

Ans. 1

Q38. If sin A = , find value of cos A.

Ans. sinA = ,

Using pythagoras theorem, H2 = P2 + B2, where H is hypoteneus, P is perpendicular and B is base.

42 = B2 + 32

B2 = 16 – 9 = 7

B =

Now,

Q39. If cos A = ,  find the value of tan A.

Ans. cosA = ,

Using Pythagoras theorem, H2 = P2 + B2, where H is hypotenuse, P is perpendicular and B is base.

132 = P2 + 122

P2 = 169 – 144 = 25

P = 5

Now, tanA =

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