Q1: What is the maximum value of
Ans: ∵ = 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 is 3.
Q2: If sin θ = 1/3, then find the value of 9 cot^{2}θ + 9.
Ans: Thus, the value is 81.
Q3: If 4 tan θ = 3, then find the value of
Ans:
Given $4\; \backslash tan\; \backslash theta\; =\; 3$4tanθ=3, we have:
Let sinθ=3k and cosθ=4k (since $\backslash tan\; \backslash theta\; =\; \backslash frac\{\backslash sin\; \backslash theta\}\{\backslash cos\; \backslash theta\}$tanθ = sinθ/cosθ). Using the identity sin^{2}θ + cos^{2}θ=1:
(3k)^{2 }+ (4k)^{2 }= $19k^2\; +\; 16k^2\; =\; 1$
9k^{2 }+ 16k^{2 }= 1
25k^{2 }= 1
⇒k= 1/5
So, sinθ = 3/5 and $\backslash cos\; \backslash theta\; =\; \backslash frac\{4\}\{5\}$cosθ = 4/5. Now calculate the expression: $\backslash tan\; \backslash theta\; =\; \backslash frac\{3\}\{$
Q4: If sin α = 1/2 and cos β = 1/2 then find the value of (α + β).
Ans:
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
⇒ sin^{2} θ + cos^{2} θ + 2 sin θ . cos θ = 3
⇒ 1 + 2 sin θ . cos θ = 3 ⇒ 2 sin θ . cos θ = 2 [∴ sin^{2}θ + cos^{2}θ = 1]
⇒ sin θ . cos θ =1 ⇒ 1 =
⇒
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:
...(1)
...(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 sec^{2}A + cosec^{2}A = sec^{2}A. csec^{2} A
Ans:
We start with the LHS:Taking a common denominator:
= sec^{2}A⋅csec^{2}A
Thus, the equation is proved.$\backslash sec^2\; A\; +\; \backslash csc^2\; A\; =\; \backslash frac\{1\}\{\backslash cos^2\; A\}\; +\; \backslash frac\{1\}\{\backslash sin^2\; A\}$
Q8: Prove that (sinθ + cosecθ)^{2} + (cosθ + secθ)^{2} = tan^{2}θ + cot^{2}θ + 7
Ans:
LHS = sin^{2}θ + cosec^{2}θ + 2sinθcosecθ + cos^{2}θ + sec^{2}θ + 2cosθsecθ
1 + 1 + cot^{2}θ + 1 + tan^{2}θ + 2 + 2
= 7 + tan^{2}θ+cot^{2}θ
Q9: If cos(40^{o} + x) = sin 30^{o}, find the value of x, provided 40o + x is an acute angle.
Ans: Given that
cos(40^{o} + x) = sin 30^{o}
Now RHS = sin 30^{o}^{ }= 1/2
So, cos (40^{o} + x) = 1/2
We know that cos60^{o}^{ } = 1/2, therefore
40 + x = 60^{o}
x = 20^{o}
Q10: Prove that .
Ans:
LHS =
(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
AB^{2} = BC^{2} + CA^{2} [Using Pythagoras Theorem]
= (2.5)^{2} + (6)^{2}
= 6.25 + 36 = 42.25
⇒
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 $\backslash al$α.
(ii) How far from the foot of the pole should he place the foot of the ladder? (Use $\backslash sqrt\{3\}\; =\; 1.73$3=1.73)
(iii) In the above situation, find the value of .
(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 $\backslash tan\; \backslash alpha$tanα:We know that tan60º=3, so the angle α=60º.
(ii) Raj needs to reach a height of 3.7m (since $51.3=3.7\hspace{0.17em}m5\; \; 1.3\; =\; 3.7\; \backslash ,\; \backslash text\{m\}$) and the ladder is inclined at 60º.
Using tan60º = 3.7/x and knowing tan60º = 1.73, we get:
Thus, the distance from the foot of the pole to the foot of the ladder is 2.14 m.
(iii) The expressionis the sine subtraction identity:
Simplifying Thus, the expression becomes:
Given α=60º, we have:
(iv) We are given BD=3 cm and BC=6 cm. Using the ratio of sides, we find that
=
This corresponds to $\mathrm{tan}30$º, so $\backslash alpha\; =\; 30^\backslash circ$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 wellbeing of the whole society.
$\backslash cos\; 60^\backslash circ\; =\; \backslash frac\{\backslash text\{adjacent\}\}\{\backslash text\{hypotenuse\}\}\; =\; \backslash frac\{x\}\{5\}$
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1. What are the basic trigonometric ratios and how are they defined? 
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5. How do you solve basic trigonometric equations? 

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