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 sin^{2}A + cos^{2}A = 1. Therefore, 1 – cos^{2}A = sin^{2}A

LHS = RHS

Hence proved.

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

**Ans**. LHS

=

Using a^{3} – b^{3} = (a – b)(a^{2} + b^{2} + ab)

=

RHS

1 + secθ cosecθ =

LHS = RHS

Hence proved.

**Q3. Prove that : (sin A + cosec A) ^{2} + (cos A + sec A)^{2} = 7 + tan^{2}A + cot^{2}A**

**Ans. **LHS

(sin A + cosec A)^{2} + (cos A + sec A)^{2} = sin^{2}A + cosec^{2}A + 2sinAcosecA + cos^{2}A + sec^{2}A + 2cosAsecA

[As we know sin^{2}A + cos^{2}A = 1, 1 + tan^{2}A = sec^{2}A, 1 + cot^{2}A = cosec^{2}A, ]

= sin^{2}A + cos^{2}A + cosec^{2}A + 2 + + sec^{2}A + 2

= 1 + 1 + cot^{2}A + 1 + tan^{2}A + 4

= 7 + tan^{2}A + cot^{2}A

LHS = RHS

Hence proved.

**Q4. Prove that **** , using the identity cosec ^{2}A- cot^{2}A = 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, H^{2} = P^{2} + B^{2}

5^{2} = 4^{2} + B^{2}

B^{2} = 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 + tan ^{2} θ = ___________.**

**Ans**. sec^{2}θ

**Q18. Find the value of **

**Ans**. = =

**Q****19****. Prove that : **

**Ans**. LHS

= = RHS

LHS = RHS

Hence proved.

**Q2****0****. 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.

**Q2****1****. 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,

OP^{2} + PQ^{2} = OQ^{2}

7^{2} + x^{2} = (x+1)^{2}

49 + x^{2} = x^{2} + 1 + 2x

48 = 2x

x = 24 cm

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

Now, SinQ =

CosQ =

**Q2****2****. If sin(A-B) = **, **cos(A+B)** **=** **, 0° < A + B ≤ 90°, A > B, find A and B.**

**Ans**. As we know

sin30^{o} =

Therefore, A – B = 30 —- (i)

Similarly, cos60^{o} =

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}

**Q2****3****. 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,

PQ^{2} + QR^{2} = PR^{2}

5^{2} + (25 – x)^{2} = x^{2}

25 + 625 – 50x + x^{2} = x^{2}

650 = 50x

x = 13 cm

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

SinP =

**Q2****4****. “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.

**Q2****5****. “sinθ= cosθ for all values of θ” (true / false)**

**Ans**. False.

As sin0^{o} ≠ cos0^{o}

**Q2****6****. “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.

**Q2****7****. “cotA is not defined for A = 0° ” (true / false)**

**Ans**. True.

As cotA = = not defined.

**Q****28****. “The value of secA is always lies between -1 and 1” (true / false)**

**Ans**. False.

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

**Q****29****. “cosec A is the product of cosec and A” (true / false)**

**Ans**. False.

**Q3****0****. “The value of cosecθ lies** ** ≥ 1 and ≤ **–**1 ” (true / false)**

**Ans.** True.

**Q3****1****. “sinθ = for some angle θ ” (true / false)**

**Ans**. False.

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

**Q3****2****. Evaluate: **

**Ans**.

[Using, sin^{2}θ + cos^{2}θ = 1]

**Q3****3****. Evaluate: **

**Ans**.

=

**Q3****4****. If sec θ =** **,** **then find sin θ.**

**Ans**. secθ = ,

Using pythagoras theorem, H^{2} = P^{2} + B^{2}, where H is hypoteneus, P is perpendicular and B is base.

13^{2} = P^{2} + 12^{2}

P^{2} = 169 – 144 = 25

P = 5

Now, sinθ =

**Q3****5****. The value of 2tan ^{2} 45° + cos^{2} 30° – sin^{2} 60° is**

**Ans**.

=

= 2

**Q****36****. sin 60° cos 30° + sin 30° cos 60° is equal to _________.**

**Ans**.

=

**Q****37****. The value of cos ^{2} θ + sin^{2} θ = __________.**

**Ans**. 1

**Q****38****. If sin A =** **, find value of cos A.**

**Ans**. sinA = ,

Using pythagoras theorem, H^{2} = P^{2} + B^{2}, where H is hypoteneus, P is perpendicular and B is base.

4^{2} = B^{2} + 3^{2}

B^{2} = 16 – 9 = 7

B =

Now,

**Q****39****. If cos A = **, ** find the value of tan A.**

**Ans**. cosA = ,

Using Pythagoras theorem, H^{2} = P^{2} + B^{2}, where H is hypotenuse, P is perpendicular and B is base.

13^{2} = P^{2} + 12^{2}

P^{2} = 169 – 144 = 25

P = 5

Now, tanA =

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