Trigonometry
Questions
Q1.
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
Q2.
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
Q3.
tan10o . tan15o . tan75o . tan80o = _____________.
(a) 4 (b) 3 (c) 2
(d) 1
Solution.
tan10o = tan(90 - 80) = cot 80o
∴ tanθ . cotθ = 1
So = tan10o .
tan15o . tan75o . tano
= cot80o . cot75o . tan75o .
tan80o
= 1
Solution
Q4.
2/3 cosec258o - 2/3 cot58o . tan 32o - 5/3 tan13o . tan 37o . tan 45o .tan53o
. tan77o
(a) 2 (b) -2 (c) -1 (d) 1
Solution.
2/3(cosec258 - cot258) - 5/3(cot77o . cot53o . tan45o . tan53o . tan77o )
= 2/3(1) - 5/3(1)
2/3 - 5/3 = -1
∴ tan(90o - 58o) = cot58o
tan(90o - θ) = cotθ
tanθ . cotθ = 1
Solution
Q5.
If θ is an acute angle and tanθ + cotθ = 2, find the value of ( tanθ + cotθ)2
(a) 5 (b) 4 (c) 2 (d) 3
Solution.
On solving tanθ + cotθ = 2
θ = 45o
= tan2θ + cot2θ + 2 tanθ . cotθ
= 2 + 2 = 4
Solution
Q6.
Find the value of θ
√3 tan2 θ - 3 = 0
(a) 60o (b) 50o
(c) 30o (d) 40o
Solution.
tan2 θ = 3/√3 = √3 = tan60o
2 θ = 60o
θ = 30o
Solution
Q7.
cos A + coa2 A = 1, then sin2A + sin4 A = 1
(a) 1 (b) 0 (c) -1 (d) 2
Solution.
(a) 1
Solution
Q8.
7sin2θ + 3cos2θ = 4
so,find the value of cotθ.
(a) 1 (b) √3
(c) 1/√3 (d) 2
Solution.
(b) √3
Solution
Q9.
If 4cotA = 5, so find the value of
5sinA - 3cosA/sinA + 2cosA
(a) 5/14 (b) 1/14 (c) 4/14 (d) 6/11
Solution.
(a) 5/14
Solution
Q10.
tan(x + y) . tan(x - y) = 1, so find the value of tan(4x/3).
(a) 2/√3 (b) √3 (c) 1 (d) 2
Solution.
(b) √3
tan(x + y) = 1/tan(x - y) = cot(x - y)
Solution
