Q1. 
Solution
Q2. Prove that,sin 50o - sin 70o + sin 10o = 0
Solution
Q3. 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


Q4. 
Solution

Q5. Prove that,
=

Solution
L.H.S.=
=
=
=
=
=
Q6. Prove that,

=
Solution
L.H.S. =
=
+
=
=
=
R.H.S.
Hence proved.
Q7. 

Solution
Q8. 
Solution
Q9. Prove that,


Solution
L.H.S.=
=
=
=
Hence proved.
=
Hence proved.
Q10. Evaluate sin105o + cos105o.
Solution
sin105o + cos105o = sin (60o + 45o) + cos (60o + 45o)
= sin60o cos45o + cos60o sin45o + cos60o cos45o - sin60o sin45o












Q11. Prove that,Cos (A + B). cos (A - B) = cos2 A - sin2 B
Solution
Q12. 
Solution
Q13. Prove that, 
Solution
Q14. If
, show that 
Solution
hence proved
Q15. 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-

Q16. 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-

Q17. Solve 
Solution
Q18. Solve 
Solution
Q19. Prove that,
=
Solution
LHS=
=
=
=
=
=
=RHS
=
Q20. 
Solution
Q21.
=
Solution
L.H.S.=
+
=
=
=
=
Q22. Convert
into radian measure.
Solution
As
Therefore
=
1o =
radians
=
radian =
radian=

Q23. 

Solution
Hence the value of 
Q24. Given 2A,2B,2C are angles of a triangle,
Prove that

Solution
Q25. 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
Q26. What is the value of

Solution
Q27. Prove that,
=
Solution
L.H.S.
=
=
=
=
R.H.S.
Hence proved.
Q28. 
Solution
Q29. Solve
Solution
Q30.
=
Solution
L.H.S.=
=
=
=
=
=
=
=
=R.H.S.
Hence proved.
Q31. Prove that,
=
Solution
L.H.S.=
dividing the expression by
=
=
=
=
=
=R.H.S.
Hence proved.
Q32. Draw the graph of y = cos2x, 


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

Q33. If
find the value of 
Solution
Using tan(x+y) we have
=
,
=
=
=
Substituting the values we get,
=
=2
=
=2
Q34. Prove that,
Solution
L.H.S.=
=
=
=
=
=
=
=
=R.H.S.
=
=
Q35. Solve 
Solution
Q36. 
Solution
Q37. 
Solution
Q38. What is the value of the supplementary angle to the sum of
.
Solution

Q39. 
Solution
Q40. Evaluate: 
Solution
Q41. 

Solution
Q42. Prove that,
Solution
L.H.S.=
=
=
[ using the property
and
]
=
=R.H.S.
H ence proved.
Q43. 
Solution

Q44. 

Solution
=
=
=
=
=
=
=R.H.S.
Hence proved.
Q45. Prove that,

Solution
L.H.S. =
=
=
=
=RHS
Q46. 
Solution
Q47. 
Solution

Q48. 
Solution
Q49. Solve 
Solution
or
Q50. 
Solution
Q51. Calculate cos15o
Solution
cos15o = cos(45o – 30o)
= cos45ocos30o + sin45o sin30o

Q52. 

Solution
Q53. Convert 22.22c into degree measure.
Solution
We know that
1c =
So degree measure =
Radian
Degree measure =
=
hence 
hence
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