Unit-5:Trigonometry
(a) Trigonometric ratios and trigonometric identities
(b)Application of trigonometry
Q1.
If 2cotA = 1, find the value of
1 - tan2A/1 + tan2A
.
(a) 5/3 (b) 6/5 (c) 4/5 (d) 3/5
Solution.
cotA =
1/2
So tanA = 2
=
1 - (2)2/1 + (2)2
=
1 - 4/1 + 4
=
3/5
Solution
Q2.
The value of tanθ is always less than 1.
(a) true (b) false (c) can't say
Solution.
(b) false
Solution
Q3.
If the angle remains the same but lengths of the sides of the triangle is varying then trigonometric ratios will _________
(a) Decrease (b) Increase (c) Remains same (d) None of these
Solution.
(c) Remains same
Solution
Q4.
The value of sinθ increase as θ increases.
(a) True (b) False (c) Constant (d) None of these
Solution.
(a) True
Solution
Q5.
Find the value of x.
tan9x = sin 45o . cos 45o + sin 30o
(a) 5o (b) 10o (c) 15o (d) 20o
Solution.
tan9x = sin 45
o . cos 45
o + sin 30
o tan 9x =
1/√2
.
1/√2
+
1/2
=
1/2
+
1/2
= 1
tan9x = 1
tan9x = tan45
o 9x = 45
o x = 5
o
Solution
Q6.
In a triangle of ABC right angled at C and ∠A = ∠B, Is tanA = tanB ?
(a) Yes (b) No (c) Can't say
Solution.
(a) Yes
Solution
Q7.
Given that cotA =
1/√3
, find the value of
= (1/cosecA) + cosA/(1/cosecA) - cosA
.
(a) 2-√3 (b) 2 (c) √3 (d) 2 + √3
Solution.
=
sinA + cosA/sinA - cosA
=
tanA + 1/tanA - 1
=
√3 + 1/√3 - 1
=
√3 + 1/√3 - 1
x
√3 + 1/√3 + 1
(√3 + 1)2/2
3 + 1 + 2√3/2
=
4 + 2√3/2
= 2 + √3
Solution
Q8.
If tanθ +
1/tanθ
= 3, find the value of
tan2θ + 1/tan2θ
(a) 3 (b) 7 (c) 6 (d) 5
Solution.
(tanθ +
1/tanθ
)
2 =
tan
2θ +
1/tan2θ
+ 2 . tanθ .
1/tanθ
9 = tan
2θ +
1/tan2θ
+ 2
tan
2θ +
1/tan2θ
= 7
Solution
Q9.
If tanθ + cotθ = 2, find the value of tan2θ + cot2θ.
(a) 4 (b) 1 (c) 3 (d) 2
Solution.
(tanθ + cotθ)2 = 4
tan2θ + cot2θ + 2.tanθ.cotθ = 4
tan2θ + cot2θ = 2
Solution
Q10.
Find the value of 3tan2A - 3(1/cosA)2.
(a) 2 (b) -3 (c) -2 (d) 3
Solution.
3tan2A - 3sec2A
3(tan2A - sec2A )
= 3(-1) -3
Solution
