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CBSE CLASS X Mathematics
Trignometry
Two mark questions with answers
Q1. Prove the identity
cos/(1 + sin) = (1 - sin)/cos
Ans1. L.H.S. = cos/(1 + sin)
= {cos/(1 + sin)} x {(1 - sin)/(1 - sin)}
= {cos(1 - sin)}/(1 - sin2)
= cos(1 - sin)/cos2
= (1 - sin)/cos = R.H.S.
Q2. Prove that cot + tan = cosec.sec
Ans2. L.H.S. = cot + tan
= (cos/sin) + (sin/cos)
= (cos2 + sin2)/(sin.cos)
= 1/(sin.cos).......[... cos2 + sin2 = 1]
= cosec.sec = R.H.S.
Q3. Prove that (cos2/sin) + sin = cosec
Ans3. L.H.S. = (cos2/sin) + sin
= (cos2 + sin2)/sin.................[... cos2 + sin2 = 1]
= 1/sin = cosec = R.H.S.
Q4. Prove the identity
{1/(1 + sinA)} + {1/(1 - sinA)} = 2sec2A
Ans4. {1/(1 + sinA)} + {1/(1 - sinA)}
= (1 - sinA + 1 + sinA)/(1 - sin2A)
= 2/cos2A..........[... cos2A + sin2A = 1]
= 2sec2A = R.H.S.
Q5. Prove that (sec - 1)/(sec + 1) = (1 - cos)/(1 + cos)
Ans5. L.H.S. = (sec - 1)/(sec + 1)
= {(1/cos) - 1}/{(1/cos) + 1}
= {(1 - cos)/cos}/{(1 + cos)/cos}
= (1 - cos)/(1 + cos) = R.H.S.
Q6. Prove that (1 + cos)/sin2 = 1/(1 - cos).
Ans6. L.H.S. = (1 + cos)/sin2
= {(1 + cos)/sin2} x {(1 - cos)/(1 - cos)}
= (1 - cos2)/{sin2(1 - cos)}
= sin2/{sin2(1 - cos)}...................[... cos2 + sin2 = 1]
= 1/(1 - cos)
= R.H.S.
Q7. Prove the identity
cos2 + cos2.cot2 = cot2
Ans7. L.H.S. = cos2 + cos2.cot2
= cos2(1 + cot2)
= cosec2cos2...............[... 1 + cot2 = cosec2]
= cos2(1/sin2)= cot2 = R.H.S.
Q8. Prove that sin2(1 + cot2) = 1
Ans8. L.H.S. = sin2(1 + cot2)
= sin2.cosec2........................[... 1 + cot2 = cosec2]
= sin2(1/sin2) = 1 = R.H.S.
Q9. Simplify cos2A + {1/(1 + cot2A)}.
Ans9. cos2A + {1/(1 + cot2A)}
= cos2A + (1/cosec2A)...........[... 1 + cot2A = cosec2A]
= cos2A + sin2A
=1
Hence cos2A + {1/(1 + cot2A)} = 1
Q10. Prove that sec2 + cosec2 = sec2cosec2.
Ans10. L.H.S. = sec2 + cosec2
= (1/cos2) + (1/sin2)
= (sin2 + cos2)/(sin2cos2)
= 1/cos2sin2
= sec2cosec2 = R.H.S.
Q11.Simplify (1 + tan2A)(1 - sinA)(1 + sinA).
Ans11. (1 + tan2A)(1 - sinA)(1 + sinA)
= (sec2A)(1 - sin2A)............[... 1 + tan2A = sec2A and sin2A + cos2A = 1]
= sec2A.cos2A
= (1/cos2A)cos2A = 1
Hence (1 + tan2A)(1 - sinA)(1 + sinA) = 1
Q12. Prove that cot - tan = (2cos2 - 1)/(sin.cos).
Ans12. L.H.S. = cot - tan
= (cos/sin) - (sin/cos) = (cos2 - sin2)/(sin.cos)
= [cos2 - (1 - cos2)]/(sin.cos).......[... sin2 + cos2 = 1]
= (2cos2 - 1)/(sin.cos) = R.H.S.
Q13. Prove that tan2 - sin2 = tan2sin2.
Ans13. L.H.S. = tan2 - sin2
= (sin2/cos2) - sin2 = sin2[(1/cos2) - 1]
= sin2{(1 - cos2)/cos2}.........[... sin2 + cos2 = 1]
= (sin2.sin2)/cos2
= tan2.sin2 = R.H.S.
Q14. Prove that (sin2A.cot2A) + (cos2A.tan2A) = 1.
Ans14. L.H.S. = sin2A.cot2A + cos2A.tan2A
= sin2A.(cos2A/sin2A) + cos2A.(sin2A/cos2A)
= cos2A + sin2A
= 1 = R.H.S.
Q15. Prove that identity sinA/(1 - cosA) = (1 + cosA)/sinA.
Ans15. L.H.S. = sinA/(1 - cosA)
= {sinA/(1 - cosA)} x {(1 + cosA)/(1 + cosA)}
= {sinA(1 + cosA)/(1 - cos2A)}..........[... sin2A + cos2A = 1]
= sinA(1 + cosA)/sin2A
= (1 + cosA)/sinA = R.H.S.
Four mark questions with answers
Q1. Prove that
Ans1.
=
=
=0
= R.H.S
Q2. Prove that
Ans2.
= cosA + sinA
= R.H.S
Q3. Prove that
Ans3.
.........[... sin2 + cos2 = 1]
= (sec + 1)/(sec - 1)
= R.H.S
Q4. Prove that
Ans4.
............[... 1 + tan2 = sec2, 1 + cot2 = cosec2]
=
=
=
=
=
(sec2 .cosec2)
(1/cos2).(1/sin2)
1/(1 - sin2)sin2..........[... sin2 + cos2 = 1]
1/(sin2  - sin4
R.H.S
Q5. Prove that (sin + cosec)2 + (cos + sec)2 = 7 + tan2 + cot2.
Ans5. L.H.S = (sin + cosec)2 + (cos + sec)2
sin2 + cosec2 + 2sin.cosec + cos2 + sec2 + 2cos.sec
= (sin2 + cos2) + cosec2 + sec2 + 2sin.(1/sin) + 2cos.(1/cos)
= 1 + cosec2 + sec2 + 2 + 2...........[... sin2 + cos2 = 1]
= 5 + 1 + cot2 + 1 + tan2
= 7 + cot2 + tan2
= R.H.S.
Q6. Prove that
Ans6.
= [(sin3A - cos3A)/{sinA.cosA(sinA - cosA)}]
= [(sinA - cosA)(sin2A + cos2A + sinA.cosA)/(sinA - cosA)sinA cosA]............[... a3 b3 = (a - b)(a2 + b2 + ab)]
= (sin2A + cos2A + sinA.cosA)/(sinA.cosA)
= (1 + sinA.cosA)/(sinA.cosA)...........[... sin2A + cos2A = 1]
= (1/sinA.cosA) + 1
= secA.cosecA + 1 = R.H.S
Q7. Prove that
2(sin6 + cos6) - 3(sin4 + cos4) + 1 = 0
Ans7. L.H.S = 2(sin6 + cos6) - 3(sin4 + cos4) +
= 2[(sin2 + cos2)3 - 3sin2.cos2(sin2 + cos2) - 3[(sin2 + cos2)2 - 2sin2.cos2]
+ 1.......[... a3 + b3 = (a + b)3 - 3ab(a + b) & a2 + b2 = (a + b)2 - 2ab]
= 2[12 - 3sin2.cos2.1] - 3[12 - 2sin2.cosec2 + 1)......[... sin2 + cos2 = 1]
= 2 - 6sin2.cos2 - 3 + 6sin2.cos2 + 1
= 2 - 3 + 1 = 0 = R.H.S
Q8. Show that
Ans8. L.H.S = [{(1)/(cosec - cot)} - {(1/sin)}]
=
= [{(sin)/(1 - cos)} - (1/sin)]
= [(sin2 - 1 + cos)/{sin(1 - cos)}]
= [cos - (1 - sin2)]/[sin(1 - cos)]
= {cos - cos2}/{sin(1 - cos)}....[... sin2 + cos2 = 1]
= {cos(1 - cos)}/{sin(1 - cos)}
= cos/sin = cot ..............(i)
R.H.S. = (1/sin) - {1/(cosec + cot)}
= (1/sin) - [1/{(1/sin) + (cos/sin)}]
= (1/sin) - {sin/(1 + cos)}
= (1 + cos - sin2)/{sin(1 + cos)}
= (cos2 + cos)/{sin(1 + cos)}.......[... sin2 + cos2 = 1]
= {cos(1 + cos)}/{sin(1 + cos)}
= cos/sin = cot ................(ii)
From (i) and (ii)
L.H.S. = R.H.S.
Q9. Prove that (tanA + secA - 1)/(tanA - secA + 1) = (1 + sinA)/cosA.
Ans9. L.H.S. = (tanA + secA - 1)/(tanA - secA + 1)
= {tanA + secA - (sec2A - tan2A)}/(tanA - secA + 1).....[Putting 1 = sec2A - tan2A as
1 + tan2A = sec2A]
= {tanA + secA - (secA + tanA)(secA - tanA)}/(tanA - secA + 1)
= {(secA + tanA) (1 - secA + tanA)}/(tanA - secA + 1)
= secA + tanA
= (1/cosA) + sinA/cosA) = (1 + sinA)/cosA = R.H.S.
Q10. Prove that
+
= 2cosec.
Ans10. L.H.S.
=
=
+
+
=
+
= [(1 + cos)/sin] + [(1 - cos)/sin].....[... sin2 + cos2 = 1]
= (1 + cos + 1 - cos)/sin
= 2/sin
= 2cosec = R.H.S.
Hence the required result.
Q11. Prove that tan - cot = (2sin2 - 1)/(sin.cos)
Ans11. L.H.S. = tan - cot
= (sin/cos) - (cos/sin)
= (sin2 - cos2)/(sincos)
= (sin2 - 1 + sin2)/(sin.cos)...........[... sin2 + cos2 = 1]
= (2sin2 - 1)/(sin.cos)
= R.H.S.
Q12. Prove that sec4 - sec2 = tan4 + tan2.
Ans12. L.H.S. sec4 - sec2
= sec2[sec2 - 1]
= (1 + tan2)[tan2].............[... sec2 - tan2 = 1]
= tan2 + tan4
= R.H.S.
Six mark questions with answers
Q1. If cot = n cot and cos = mcos, then prove that sin2 = (m2 - 1)/(n2 - 1).
Ans1. cot = n cot
 cot = cot/n
 tan = n/cot..............(1)
cos = mcos
 cos = cos/m
 sec = m/cos..............(2)
Squaring (1) and (2) and subtracting, we get
tan2 - sec2 = (n2/cot2) - (m2/cos2)
 -1 = (n2sin2/cos2) - (m2/cos2)......[... sec2 = 1 + tan2]
 (m2/cos2) - (n2sin2/cos2) = 1
 (1/cos2)[m2 - n2sin2] = 1
 m2 - n2sin2 = cos2
 m2 - n2sin2 = 1 - sin2.....[... sin2 + cos2 = 1]
 m2 - 1 = n2sin2 - sin2
 m2 - 1 = (n2 - 1)sin2
 sin2 = (m2 - 1)/(n2 - 1)
Hence Proved.
Q2. The angle of elevation of a tower at a point is 45o. After going 40 m towards the
foot of the tower, the angle of elevation of the tower becomes 60o. Find the height of
the tower.
Ans2.
Let AB = h metres be the height of the
tower. Let C & D be the points where the
angles subtended are 450 and 60o.
 CD = 40 m
Let DB = x m and BC = CD + DB
= (40 + x)m
In  ABD
tan 60o = AB/BD
 3 = h/x
 h = 3x
 x = h/3.................(1)
In  ABC
tan45o = AB/BC
 1 = h/(40 + x)
 h = 40 + x
Putting the value of x from (1), we get
h = 40 + h/3
 h - (h/3) = 40
 h(1 - 1/3) = 40
 h{(3 - 1)/3} = 40
 h = 403/(3 - 1)
= 403/(3 - 1) x {(3 + 1)/(3 + 1)}
= 40(3 + 3)/(3 - 1) = 40(3 + 3)/2
= 20(3 + 1.732) = 20 x 4.732 = 94.64 m
Hence, the height of the tower is 94.64 m.
Q3. From the top of a building 60 m high, the angles of depression of the top and
bottom of a tower are observed to be 30o and 60o. Find the height of the tower.
Ans3.
Let h be the height of the tower
AB = 60 m
Suppose BD = t m
In  ABD
tan 60o = AB/BD
3 = 60/t
 t = 60/3................(1)
In  AEC
tan 30o = AE/CE
 tan 30o = AE/BD..................[... CE = BD]
 1/3 = AE/t
 1/3 = AE/(60/3)........[Putting the value of t from (1)]
 AE = (1/3) x (60/3)
 AE = 20 m
Now h = CD = EB
= AB - AE
= (60 - 20)m
= 40 m
Thus, the height of the tower = 40 m.
Q4. The horizontal distance between two towers is 80 m. The angle of depression of
the top of first tower when seen from the top of the second tower is 35 o. If the
height of the second tower is 160 m, find the height of the first tower.
Ans4.
Let AB be the tower 160 m high.
Let h be the height of the other tower DE
Now BD = 80 m
EAC = 35o
 AEF = 35o
In  AEF
tan35o = AF/EF
 0.7002 = (AB - FB)/EF
 0.7002 = (160 - h)/80
 0.7002 x 80 = 160 - h
 56.016 = 160 - h
 h = 160 - 56.016
 h = 103.984
 Height of the tower = 103.984 m.