Q1. 
Solution

Q2. 
Solution
Q3.
=
Solution
L.H.S.=
+
=
=
=
=
Q4. What is the value of

Solution
Q5. 
Solution

Q6. Calculate cos15o
Solution
cos15o = cos(45o – 30o)
= cos45ocos30o + sin45o sin30o

Q7. 
Solution
Q8. If
find the value of 
Solution
Using tan(x+y) we have
=
,
=
=
=
Substituting the values we get,
=
=2
=
=2
Q9. 
Solution
Q10. Prove that,
=
Solution
L.H.S.=
dividing the expression by
=
=
=
=
=
=R.H.S.
Hence proved.
Q11. Prove that,


Solution
L.H.S.=
=
=
=
Hence proved.
=
Hence proved.
Q12. Prove that,
Solution
L.H.S.=
=
=
[ using the property
and
]
=
=R.H.S.
H ence proved.
Q13. Convert 22.22c into degree measure.
Solution
We know that
1c =
So degree measure =
Radian
Degree measure =
=
hence 
hence
Q14. 

Solution
Q15. Prove that,Cos (A + B). cos (A - B) = cos2 A - sin2 B
Solution
Q16. 
Solution
Q17. Solve 
Solution
Q18. Evaluate: 
Solution
Q19. Prove that, 
Solution
Q20. 
Solution
Q21. Prove that,
=
Solution
LHS=
=
=
=
=
=
=RHS
=
Q22. Prove that,
Solution
L.H.S.=
=
=
=
=
=
=
=
=R.H.S.
=
=
Q23. Prove that,
=

Solution
L.H.S.=
=
=
=
=
=
Q24. Prove that,

Solution
L.H.S. =
=
=
=
=RHS
Q25. 
Solution

Q26. Given 2A,2B,2C are angles of a triangle,
Prove that

Solution
Q27. Draw the graph of y = cos2x, 


Solution
So, the graph of above function is as follows-

Q28. 

Solution
Q29. Prove that,sin 50o - sin 70o + sin 10o = 0
Solution
Q30. Draw the graph of y = 
Solution
To draw the graph, we shall first find the values of x in the interval [0,
].
We know that,
Hence the graph of the given function is as follows-

Hence the graph of the given function is as follows-

Q31. Solve 
Solution
Q32. 
Solution
Q33. 

Solution
Hence the value of 
Q34. 
Solution
Q35. Solve 
Solution
Q36. Convert
into radian measure.
Solution
As
Therefore
=
1o =
radians
=
radian =
radian=

Q37. 
Solution
Q38. 
Solution
Q39. Evaluate sin105o + cos105o.
Solution
sin105o + cos105o = sin (60o + 45o) + cos (60o + 45o)
= sin60o cos45o + cos60o sin45o + cos60o cos45o - sin60o sin45o












Q40. 
Solution
Q41. Find the angle between the minute hand and the hour hand of a clock when the time is 5:20.
Solution
In 12 hours hour hand moves by 360 degrees
In 1 hr it will move by 30 degrees
So in 1 min it will move ½ degree
So in 5 hr and 20 min it will move 5 x 30 + 20 x ½ = 160 degree
For minute hand
In 60 min it moves 360 degree
So in 1 min it moves 6 degree.
So in 20 min it moves 20x6 =120 degree.
Therefore difference is 160 – 120 = 40 degree
Q42.
=
Solution
L.H.S.=
=
=
=
=
=
=
=
=R.H.S.
Hence proved.
Q43. If
, show that 
Solution
hence proved
Q44. What is the value of the supplementary angle to the sum of
.
Solution

Q45. Solve
Solution
Q46. 
Solution
Q47. Prove that,

=
Solution
L.H.S. =
=
+
=
=
=
R.H.S.
Hence proved.
Q48. Solve 
Solution
or
Q49. 

Solution
=
=
=
=
=
=
=R.H.S.
Hence proved.
Q50. The speed of sound is nearly 341m/s, if a car of wheel radius 70 cm moves with a speed of sound. Find the number of revolutions it will make in 10 sec.
Solution
Let n be the number of revolution a wheel makes in 10 sec.
The distance travelled by the wheel in 10 sec is 3410 m
Therefore, n x
= 3410


Q51. Draw the graph of y = tanx
Solution
We know that the value of tanx repeats after an interval of
. To draw the graph we shall first find the values of tanx for some convenient values of x in [0,
]
We know that tan90o is not defined, but as the angle x approaches 90o from left side the value of tanx increases indefinitely. Similarly as the angle x approaches 90o from right side, the value of tanx decreases indefinitely.
Hence the graph of above function is as follows-

We know that tan90o is not defined, but as the angle x approaches 90o from left side the value of tanx increases indefinitely. Similarly as the angle x approaches 90o from right side, the value of tanx decreases indefinitely.
Hence the graph of above function is as follows-

Q52. Prove that,
=
Solution
L.H.S.
=
=
=
=
R.H.S.
Hence proved.
Q53. 

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