Q1: What is the maximum value of 3cosec θ ?
Ans: ∵ 3cosec θ = 3 sin θ
Since the maximum value of ‘sin θ’ is 1
∴ the maximum value of 3 sin θ is 3 × 1 i.e. 3
⇒ the maximum value of 3cosec θ is 3.
Q2: If sin θ = 1/3, then find the value of 9 cot2θ + 9.
Ans:
cot2 θ = cosec2 θ - 1
cosec2θ = 1sin2θ = 1(1/3)2 = 9
cot2 θ = 9 - 1 = 8
9 cot2 θ + 9 = 9(8) + 9 = 72 + 9 = 81
Thus, the value is 81.
Q3: If 4 tan θ = 3, then find the value of 4 sin θ - cos θ4sin θ + cos θ
Ans: Given 4 tanθ=3, we have: tan θ = 34
Let sinθ=3k and cosθ=4k (since tanθ = sinθ/cosθ). Using the identity sin2θ + cos2θ=1:
(3k)2 + (4k)2 =
9k2 + 16k2 = 1
25k2 = 1
⇒k= 1/5
So, sinθ = 3/5 and cosθ = 4/5. Now calculate the expression:
= 4 sin θ - cos θ4sin θ + cos θ
= 4 x 3/5 - 4/5 4 x 3/5 + 4/5
= 12/5 - 4/5 12/5 + 4/5
= 8/5 16/5
= 1/2
Q4: If sin α = 1/2 and cos β = 1/2 then find the value of (α + β).
Ans: sin 30º = 1/2 , sin α = 1/2 ⇒ α = 30º
cos 60º = 1/2 , cos β = 1/2 ⇒ β = 60º
⇒ ( α + β) = 30º + 60º = 90º
Q5: If sin θ + cos θ = √3 , find the value of tan θ + cot θ .
Ans: sin θ + cos θ = √3 ⇒ (sin θ + cos θ)2 = 3
⇒ sin2 θ + cos2 θ + 2 sin θ . cos θ = 3
⇒ 1 + 2 sin θ . cos θ = 3 ⇒ 2 sin θ . cos θ = 2 [∴ sin2θ + cos2θ = 1]
⇒ sin θ . cos θ =1 ⇒ 1 = 1sin θ .cosθ
⇒ 1 = sin2θ + cos2θ sin θ .cosθ
= sin θ cosθ + cos θ sin θ = tanθ + cotθ
Thus, tanθ + cotθ = 1
Q6: cos (A+B) = 1/2 and sin (A–B) = 1/2 ; 0° < (A + B) < 90° and (A – B) > 0°. What are the values of ∠A and ∠B?
Ans: We have
cos (A+B) = 1/2 ,
Also, cos 60° = 1/2 ⇒ A+B = 60° ...(1)
Also, sin (A–B) = 1/2
and sin 30° = 1/2 ⇒ A - B = 30° ...(2)
Adding (1) and (2), 2A = 90 ⇒ A = 45
From (1) 45° + B = 60° ⇒ B = 60° – 45° = 15°
Thus, ∠A = 45° and ∠B = 15°
Q7: Prove that sec2A + cosec2A = sec2A. cosec2 A
Ans: We start with the LHS:
sec2A + cosec2A = 1 cos2A + 1 sin2A
Taking a common denominator:
sin2A + cos2A sin2 A. cos2A = 1 sin2 A. cos2A
= sec2A⋅cosec2A
Thus, the equation is proved.
Q8: Prove that (sinθ + cosecθ)2 + (cosθ + secθ)2 = tan2θ + cot2θ + 7
Ans:
LHS = sin2θ + cosec2θ + 2sinθcosecθ + cos2θ + sec2θ + 2cosθsecθ
1 + 1 + cot2θ + 1 + tan2θ + 2 + 2
= 7 + tan2θ+cot2θ
Q9: If cos(40o + x) = sin 30o, find the value of x, provided 40o + x is an acute angle.
Ans: Given that
cos(40o + x) = sin 30o
Now RHS = sin 30o = 1/2
So, cos (40o + x) = 1/2
We know that cos60o = 1/2, therefore
40 + x = 60o
x = 20o
Q10: Prove that .
Ans:
LHS =
(using 1- cos2 θ = sin2θ )
= 1 + cosθ + 1 - cosθsin θ = 2 sin θ = 2 cosec θ
(i) Find the length of the ladder.
(ii) Which mathematical concept is used in this problem?
(iii) By putting up a banner in favour of respect to women, which value is depicted by the group of students?
Ans: (i) Let AB be the ladder and CA be the wall with the window at A. Also,
BC = 2.5 m
CA = 6m
∴ In rt ΔACB, we have
AB2 = BC2 + CA2 [Using Pythagoras Theorem]
= (2.5)2 + (6)2
= 6.25 + 36 = 42.25
⇒ AB = √42.25 =6.5 m
Thus, the length of the ladder is 6.5 m.
(ii) Triangles (Pythagoras Theorem)
(iii) Creating positive awareness in public regarding women.
Q2: Raj is an electrician in a village. One day power was not there in entire village and villagers called Raj to repair the fault. After thorough inspection he found an electric fault in one of the electric pole of height 5 m and he has to repair it. He needs to reach a point 1.3m below the top of the pole to undertake the repair work. On the basis of above, answer the following question.
(i) When the ladder is inclined at an angle of α such that 3tan α + 2 = 5 to the horizontal, find the angle α.
(ii) How far from the foot of the pole should he place the foot of the ladder? (Use 3=1.73)
(iii) In the above situation, find the value of sin α cos α2 - cos α sin α2
(iv) In the above situation, if BD=3 cm and BC=6 cm, find α.
(v) In your opinion, how does Raj's action reflect the values of responsibility and community service?
Ans:
(i) Given the equation 3tanα+2=5, simplify to find tanα:
√3 tan α = 5 - 2 = 3
tan α = 3√3 = √3
We know that tan60º=3, so the angle α=60º.
(ii) Raj needs to reach a height of 3.7m (since ) and the ladder is inclined at 60º.
Using tan60º = 3.7/x and knowing tan60º = 1.73, we get:
x = 3.71.73 = 2.14 m
Thus, the distance from the foot of the pole to the foot of the ladder is 2.14 m.
(iii) The expression sin α cos α2 - cos α sin α2 is the sine subtraction identity:
sin α cos α2 - cos α sin α2 = sin( α - α2 )
Simplifying α - α2 = α2
Thus, the expression becomes: sin α2
Given α=60º, we have:
sin 60º2 = sin30º = 12
(iv) We are given BD=3 cm and BC=6 cm. Using the ratio of sides, we find that tan α = BDBC
= 36 = 12
This corresponds to º, so α=30º.
(v) Raj's actions reflect a strong sense of responsibility because he promptly responded to the village's call for help, even in challenging conditions. His commitment to restoring power shows that he cares about the welfare of others. This sense of responsibility is a key value in community service, where individual efforts contribute to the well-being of the whole society.
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1. What are the basic trigonometric ratios and how are they defined? |
2. How can trigonometry be applied in real-life situations? |
3. What is the Pythagorean theorem and how does it relate to trigonometry? |
4. What are some common identities in trigonometry? |
5. How do you solve basic trigonometric equations? |
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