Chapter: Introduction to Trigonometry Subject: Mathematics (Class 10)
Chapter Practice Sheet: Introduction to Trigonometry
Questions
2 2
Q1. Given that sin θ = aa2 −b
+b2
, find the value of cos θ. Using this result, explain why the value
of sec θ must always be greater than or equal to 1 (assuming a > b > 0).
√
Q2. If 3 tan 2θ − 3 = 0, where 0◦ < 2θ < 90◦ , determine the value of the expression:
sin θ · cos θ − tan2 θ
Q3. If x = p sec θ + q tan θ and y = p tan θ + q sec θ, prove that x2 − y 2 = p2 − q 2 .
Q4. In a right-angled triangle ABC (right-angled at B), the difference between the length of
the hypotenuse AC and the base BC is 2 cm. If the hypotenuse AC is 4 cm, calculate
the measure of ∠A.
Q5. Find the acute angle θ that satisfies the equation:
2 sin2 θ − cos θ = 1
Q6. Prove that:
tan A cot A
+ = 2 sin A cos A
1 + tan A 1 + cot2 A
2
Q7. If 5 tan α = 4, evaluate the following expression without finding the value of α:
5 sin α − 3 cos α
5 sin α + 2 cos α
Q8. If sec θ + tan θ = x, prove that:
x2 − 1
sin θ =
x2 + 1
√
Q9. If sin(A + B) = 1 and cos(A − B) = 2
3
, where A > B and 0◦ < A + B ≤ 90◦ , find the
values of A and B.
Q10. Prove that: r r
1 + sin θ 1 − sin θ
+ = 2 sec θ
1 − sin θ 1 + sin θ
Page 1
, Chapter: Introduction to Trigonometry Subject: Mathematics (Class 10)
Detailed Solutions
Solution 1:
2 −b2
Given sin θ = aa2 +b 2 2 2 2
2 . Let the opposite side be k(a − b ) and the hypotenuse be k(a + b ).
By Pythagoras Theorem:
Base2 = Hypotenuse2 − Perpendicular2
= k 2 (a2 + b2 )2 − k 2 (a2 − b2 )2
= k 2 [(a4 + b4 + 2a2 b2 ) − (a4 + b4 − 2a2 b2 )]
= k 2 (4a2 b2 )
=⇒ Base = 2abk
Base 2abk 2ab
Now, cos θ = Hypotenuse
= k(a2 +b2 )
=
a2+ b2
Reasoning for sec θ:
1 a2 + b 2
= sec θ =
cos θ 2ab
Since (a − b) ≥ 0 =⇒ a + b − 2ab ≥ 0 =⇒ a + b2 ≥ 2ab. Therefore, the numerator is ≥
2 2 2 2
denominator. Hence, sec θ ≥ 1.
Solution 2:
We have the equation:
√
3 tan 2θ − 3 = 0
√
3 tan 2θ = 3
3 √
tan 2θ = √ = 3
3
√
Since tan 60◦ = 3, we equate the angles:
2θ = 60◦ =⇒ θ = 30◦
Substituting θ = 30◦ into the given expression:
Expression = sin 30◦ · cos 30◦ − tan2 30◦
√ ! 2
1 3 1
= − √
2 2 3
√
3 1
= −
4 3
√
3 3−4
=
12
Solution 3:
Given:
x = p sec θ + q tan θ and y = p tan θ + q sec θ
Page 2
Chapter Practice Sheet: Introduction to Trigonometry
Questions
2 2
Q1. Given that sin θ = aa2 −b
+b2
, find the value of cos θ. Using this result, explain why the value
of sec θ must always be greater than or equal to 1 (assuming a > b > 0).
√
Q2. If 3 tan 2θ − 3 = 0, where 0◦ < 2θ < 90◦ , determine the value of the expression:
sin θ · cos θ − tan2 θ
Q3. If x = p sec θ + q tan θ and y = p tan θ + q sec θ, prove that x2 − y 2 = p2 − q 2 .
Q4. In a right-angled triangle ABC (right-angled at B), the difference between the length of
the hypotenuse AC and the base BC is 2 cm. If the hypotenuse AC is 4 cm, calculate
the measure of ∠A.
Q5. Find the acute angle θ that satisfies the equation:
2 sin2 θ − cos θ = 1
Q6. Prove that:
tan A cot A
+ = 2 sin A cos A
1 + tan A 1 + cot2 A
2
Q7. If 5 tan α = 4, evaluate the following expression without finding the value of α:
5 sin α − 3 cos α
5 sin α + 2 cos α
Q8. If sec θ + tan θ = x, prove that:
x2 − 1
sin θ =
x2 + 1
√
Q9. If sin(A + B) = 1 and cos(A − B) = 2
3
, where A > B and 0◦ < A + B ≤ 90◦ , find the
values of A and B.
Q10. Prove that: r r
1 + sin θ 1 − sin θ
+ = 2 sec θ
1 − sin θ 1 + sin θ
Page 1
, Chapter: Introduction to Trigonometry Subject: Mathematics (Class 10)
Detailed Solutions
Solution 1:
2 −b2
Given sin θ = aa2 +b 2 2 2 2
2 . Let the opposite side be k(a − b ) and the hypotenuse be k(a + b ).
By Pythagoras Theorem:
Base2 = Hypotenuse2 − Perpendicular2
= k 2 (a2 + b2 )2 − k 2 (a2 − b2 )2
= k 2 [(a4 + b4 + 2a2 b2 ) − (a4 + b4 − 2a2 b2 )]
= k 2 (4a2 b2 )
=⇒ Base = 2abk
Base 2abk 2ab
Now, cos θ = Hypotenuse
= k(a2 +b2 )
=
a2+ b2
Reasoning for sec θ:
1 a2 + b 2
= sec θ =
cos θ 2ab
Since (a − b) ≥ 0 =⇒ a + b − 2ab ≥ 0 =⇒ a + b2 ≥ 2ab. Therefore, the numerator is ≥
2 2 2 2
denominator. Hence, sec θ ≥ 1.
Solution 2:
We have the equation:
√
3 tan 2θ − 3 = 0
√
3 tan 2θ = 3
3 √
tan 2θ = √ = 3
3
√
Since tan 60◦ = 3, we equate the angles:
2θ = 60◦ =⇒ θ = 30◦
Substituting θ = 30◦ into the given expression:
Expression = sin 30◦ · cos 30◦ − tan2 30◦
√ ! 2
1 3 1
= − √
2 2 3
√
3 1
= −
4 3
√
3 3−4
=
12
Solution 3:
Given:
x = p sec θ + q tan θ and y = p tan θ + q sec θ
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