Q1.
Find the value of sin2 41 + sin2 49
(a) 1 (b) 2 (c) 3 (d) 4
Solution.
sin(90o - 41o) = cos41o sin49o
= sin241o + cos2
41o = 1
Solution
Q2.
If sinθ + sin2θ = 1, then cos2θ + cos4θ =
(a) 2 (b) 1 (c) 3
(d) 4
Solution.
sinθ = 1 - sin2θ = cos2θ
sin2θ = cos4θ
So, cos2θ + cos4θ = sinθ + sin2θ = 1
Solution
Q3.
What is the value of 1/(1 + cot2θ) + 1/(1 + tan2θ)
(a) 4 (b) 3 (c) 1
(d) 2
Solution.
= 1/cosec2θ + 1/sec2θ = sin2θ + cos2θ = 1
Solution
Q4.
If A,B,C are the interior angles of a triangle ABC , then tan(B + C/2) =
(a) cotC (b) cot(A/2) (c) tan(A/2) (d) tan B
Solution.
In triangle ABC we know that
A + B + C = 180o
B + C = 180o - A
B + C/2 = 90 - A/2
tan(B + C/2) = tan(90 - A/2) = cotA/2
Solution
Q5.
Find the value of A , If sec4A = cosec(A - 20) where 4A is an acute angle.
(a) 72o (b) 22o (c) 33o
(d) 45o
Solution.
sec4A = sec{90 - (A - 20)}
sec4A = sec{90o - (A - 20)}
4A = 90o - A + 20o
5A = 110o
A = 22o
Solution
Q6.
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
Q7.
tan1o . tan2o . tan3o ---------tan89o
(a) 4 (b) 3 (c) 2 (d) 1
Solution.
= tan1o . tan2o . tan3o ---------tan89o
= tan(90o - 89o) .
tan(90o - 88o) tan(90o - 87o)----------tan89o
= cot89o . cot88o . cot87o . -------tan45o-----tan87o . tan88o . tan89o
= 1
∴ tanθ . cotθ = 1
Solution
Q8.
Find the value of x.
√2sin 3x = 1
(a) 20o (b) 15o (c) 10o (d) 5o
Solution.
sin 3x = 1/√2
sin 3x = sin45o
3x = 45o
x = 15o
Solution
Q9.
If tanθ + cotθ = 2, find the value of tan9θ + cot9θ .
(a) 5 (b) 4 (c) 2 (d) 3
Solution.
tanθ + 1/tanθ = 2
tan2 θ - 2 tanθ + 1 = 0
(tanθ - 1)2 = 0
tanθ = 1
tanθ = tan45o
θ = 45o
= tan9
θ + cot9θ
= (tanθ)9 + (cotθ)9
(tan45o)
9 + (cot45o)9
= 19 + 19
= 1 + 1 = 2
Solution
Q10.
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
