Download 2. - Testlabz.com

Question Bank
Trigonometry
cos3 A + sin 3 A
cos3 A – sin 3A
+
1. Prove that
=2
cos A + sinA
cos A – sinA
cos3 A + sin 3 A
cos3 A – sin 3A
Solution. L.H.S. =
+
cos A + sinA
cos A – sinA
=
(cos A + sinA) (cos 2A + sin 2A – cosA sinA) (cos A – sinA) cos 2A + sin 2 A + cosA sinA
+
(cos A + sinA)
(cos A – sinA)
=
=
=
⎡⎣∵ a 3 + b3 = (a + b) (a 2 + b 2 – ab) and a3 – b3 = (a – b) (a 2 + b 2 + ab)⎤⎦
(cos 2 A + sin 2 A – cosAsinA) + (cos 2 A + sin 2 A + cosA sinA)
(1 – cosA sinA) + (1 + cosA sinA) ⎡⎣∵ cos 2 A + sin 2 A = 1⎤⎦
= 1 – cosA sinA + 1 + cosA sinA = 2 = R.H.S. Proved.
cosA
sinA
+
2. Prove that
= cosA + sinA
1 – tanA 1 – cotA
Solution.
cosA
sinA
L.H.S. =
+
1 – tanA 1 – cotA
cosA
sinA
=
+
sinA
cosA
1–
1–
cosA
sinA
cosA × cosA
sinA × sinA
=
+
cos A – sinA
cos A – sinA
cos 2 A
sin 2 A
=
–
cos A – sinA cos A – sin A
cos 2 A – sin 2 A
=
cosA – sinA
(cosA + sinA) (cosA – sinA)
=
(cosA – sinA)
Math Class X
1
Question Bank
[∵a2 – b2 = (a + b) (a – b)]
= cosA + sinA = R.H.S. Proved.
3. Prove that =
sinA
sinA
=2+
cotA + cosecA
cotA – cosecA
Solution.
sinA
cotA + cosecA
sinA
sinA
sinA
=
=
=
cosA
1
cosA + 1 cosA + 1
+
sinA
sinA
sin A
1 – cos 2 A
⎡⎣∵ sin 2 θ = 1 – cos 2 θ ⎤⎦
=
cos A + 1
(1 + cosA)(1 – cosA)
=
= 1 – cosA
(cosA + 1)
sinA
R.H.S. = 2 +
cotA – cosecA
sinA
sin 2 A
= 2+
= 2+
cosA
1
cosA – 1
–
sinA
sinA
1 – cos 2 A
⎡⎣∵ sin 2 θ = 1 – cos 2 θ ⎤⎦
= 2+
cosA – 1
(1 + cosA) (1 – cosA)
=2+
cosA – 1
(1 + cosA) (1 – cosA)
= 2–
cosA – 1
L.H.S. =
= 2 – (1 + cosA)
= 2 – 1 – cos A = L.H.S. Proved.
Math Class X
2
Question Bank
4. If sinA + cosA = m and secA + cosecA = n, prove that n (m2 – 1) =
2 m.
Solution. We have,
m = sinA + cosA
m2 = (sinA + cosA)2
⇒
= sin2A + cos2A + 2sinA cosA
= 1 + 2 sin A cosA
2
m – 1 = 1 + 2 sinA cosA – 1
⇒
= 2 sinA cosA
2
⇒ n (m – 1) = (secA + cosecA). 2sinA cosA
= 2 sinA cosA secA + 2 sinA cosA cosecA
= 2 sinA + 2cosA
[∵cosA secA = 1 and sinA cosec A = 1]
= 2(sinA + cosA) = 2m
2
Hence, n (m – 1) = 2m. Proved.
5.
Prove that
1
1
1
1
–
=
–
(sec A + tanA)
cosA
cosA (secA – tanA)
Solution.
1
1
–
secA + tanA cosA
1
1
=
–
1
sinA
cosA
+
cos A
cosA
cosA
1
cos 2 A – 1 – sinA
=
–
=
1 + sinA cosA
cos A ( 1 + sinA)
L.H.S. =
1 – sin 2 A – 1 – sinA
sinA(1 + sinA)
=
=
cos A(1 + sinA)
cos A (1 + sinA)
= – tanA.
Math Class X
3
Question Bank
1
1
–
cosA secA – tanA
1
1
–
=
1
sinA
cosA
–
cosA cosA
1
cosA
=
–
cosA 1 – sinA
1 – sinA – cos 2 A
=
cos A (1 – sinA)
1 – sinA – 1 + sin 2 A
– sinA(1 – sinA)
=
=
cos A(1 – sinA)
cos A(1 – sinA)
=
– tanA.
Hence, LHS = RHS. Proved.
6. If x sin3θ + y cos3θ = sinθ cosθ and x sinθ – y cosθ = 0, then prove
that x2 + y2 = 1
Solution. We have
x sin3θ + y cos3θ = sinθ cosθ
... (i)
x sinθ – y cosθ = 0
... (ii)
sinθ
y
⇒
…(iii)
=
x
cosθ
sin 2 θ
cos 2 θ
+ y.
=1
From (i) x.
cosθ
sinθ
sinθ
cosθ
sinθ + y.
cosθ = 1
⇒ x.
cosθ
sinθ
y
x
⇒ x. sinθ + y. . cosθ = 1 [From (iii)]
x
y
⇒ y sinθ + x cosθ = 1
⇒ x cosθ + y sinθ = 1
... (iv)
Squaring (ii) and (iv) and adding, we get,
(x sinθ – y cosθ)2 + (x cosθ + y sinθ)2 = 0 + 12
⇒ x sin2θ – y2 cos2θ – 2xy sinθ cosθ + x2 cos2θ + yi2 sin2 θ + 2xy
sinθ cosθ = 1
R.H.S. =
Math Class X
4
Question Bank
⇒ x2 + ( sin2 θ + cos2 θ) + y2 (cos2 θ + sin2 θ) = 1
⇒ x2 + y2 = 1. Proved.
cosθ
cosθ
7. Is
+
= 2an identity? If not solve for θ,
cosecθ + 1 cosecθ – 1
where 0° < θ < 90°.
Solution.
cosθ
cosθ
Here, LHS
+
cosecθ + 1 cosecθ – 1
cosθ
cosθ
+
=
1
1
+1
–1
sin θ
sinθ
cosθ sinθ cosθ sinθ
=
+
1 + sinθ
1 – sinθ
cosθ sinθ ( 1– sinθ + 1 + sinθ)
=
1 – sin 2 θ
2sinθ cosθ
=
= 2 tanθ
cos 2 θ
Thus, the given equality becomes 2 tanθ = 2
If the equality holds true for all values of θ, then the equality is an
identity.
Let us take θ = 30°
2
So, 2 tan θ = 2 tan30° =
3
⇒ 2 tan θ ≠ 2 for θ = 30°
Therefore the equality is not an identity. It is an equation.
Now, 2 tan θ = 2 ⇒ tan θ = 1 ⇒ tan θ = tan 45° ⇒ θ = 45°.
8. If tan θ + sec θ = 3, where θ is acute, then prove that 5 sinθ = 4.
Solution. We have tanθ + secθ = 3
sinθ
1
⇒
+
=3
cosθ
cosθ
1 + sinθ
⇒
=3
cosθ
Math Class X
5
Question Bank
(1 + sinθ)2 = 9 cos2θ
[Squaring both sides]
2
1 + sin θ + 2 sinθ = 9 – 9 sin2θ
10 sin2θ + 2 sinθ – 8 = 0
10 sin2θ + 10 sinθ – 8 sinθ – 8 = 0
10 sinθ(sinθ + 1) – 8 (sinθ+ 1) = 0
(sinθ + 1) (10 sinθ – 8) = 0
8
4
⇒ sinθ= –1 or sinθ =
=
10
5
4
⇒ sin θ = [Rejecting sinθ= –1, since θis acute]
5
⇒ 5 sinθ = 4. Proved.
9. Without using trigonometric tables, prove that : tan 10º tan 20º tan
1
30º tan 70º tan 80º =
3
Solution.
L.H.S. = tan 10º tan 20º tan 30º tan 70º tan 80º
= (tan 10º tan 80º ), (tan 20º tan 70º) tan 30º
= tan (90º – 80º) tan 80º. tan (90º – 70º) tan 70º tan 30º
= cot 80º tan 80º. cot70º tan 70º tan 30º
[∵ tan (90º – θ ) = cot θ ]
1
1
=
tan 70º tan 30º
. tan80º.
tan 80º
tan 70º
1
= 1.1. tan 30º =
= R.H.S. Proved.
3
10. Prove that
sinA
cosA
= sec (90º – A ) cosec (90º – A)
+
sin(90º – A) cos(90º – A)
Solution.
sinA
cosA
L.H.S. =
+
sin(90º – A)
(cos(90º – A)
sinA
cosA
=
+
cosA
sinA
⇒
⇒
⇒
⇒
⇒
⇒
Math Class X
6
Question Bank
= [∵ sin (90º – A) = cosA and cos (90º – A) = sinA]
sin 2 A + cos 2 A
1
⎡⎣∵ sin 2 A + cos 2 A = 1⎤⎦
=
=
sin A cos A
sinA cosA
= cosecA secA
R.H.S = sec ( 90º – A ) cosec (90º – A)
=
cosecA secA
[∵ sec (90º – A) = cosec A and cosec (90º – A) = secA]
= L.H.S.
Hence, L.H.S. = R.H.S. Proved.
2
11. Prove that
2
2
sin 20º + sin 70º
2
sin (90º – θ) sinθ
+
2
cos 20º + cos 70º
tanθ
+
cos(90º – θ) cosθ
cotθ
Solution.
2
L.H.S.=
2
sin 20º + sin 70º
2
2
cos 20º + cos 70º
2
+
sin (90º – θ) sinθ
tan θ
+
cos(90º – θ) cosθ
cotθ
sin (90º – 70º ) sin 70º
cosθ sinθ sinθ cosθ
+
+
cos 2 (90º – 70º) cos 2 70º
tan θ
cotθ
cos 2 70º + sin 2 70º cosθ sinθ sinθ cosθ
=
+
+
sinθ
cosθ
sin 2 70 + cos 2 70º
cos θ
sin θ
= [∵ sin (90º – θ) = cosθ, cos (90º – θ ) = sinθ]
1
+ cos 2 θ + sin 2 θ
=
1
= 1 + 1 = 2 = R.H.S. Proved.
12. Using the tables, find the values of
(i) sin 60º 23′
(ii) cos 21º 56′ (iii) tan 75º 2′ (iv) cot 40º 36′
Solution. From trigonometric tables, we have
(i)
sin 60º 18′ = 0.8689
2
2
=
Mean difference for 5′
⇒
(ii)
sin 60º 23′
(To be added)
= 0.8696
cos 251º54′
Mean difference for 2′
Math Class X
=7
= 0.9278
= 2
(To be subtracted)
7
Question Bank
=2
⇒
21º 56′ = 0.9276
(iii)
tan
75º
= 3.7321
Mean difference for 2′
= 93
(To be added)
⇒
75º 2′ = 3.7414
(iv)
cot 40º 36′ = cot (90º – 49º 24′) = tan 49º 24′
Now, tan 49º 24′
= 1.1667
13. Find θ when
(i) sin θ = 0.0990 (ii) cos θ = 0.5536 (iii) tan θ = 5.2010
Solution.
(i) From the table, find the angle whose sine is just smaller
than 0.0990
sin θ =
0.0990
sin 5º 36′ =
0.0976
Difference = 0.0014
Mean difference 14 corresponds to 5′
∴ Required angle = (5º 36′ + 5′= 5º 41′.
(ii)
From the table, find the angle whose cosine is just greater
than 0.5536
cos θ =
sin 5º 36′
0.5536
= 0.5548
Difference = 0.0012
Mean difference 12 corresponds to 5′
∴ Required angle
= (56º 18′ + 5′) = 56º 23′.
(iii) From the table, find the angle whose tangent is just smaller than
5.2010
tan θ = 5.2010
tan 79º 6′ = 5.1923
Since mean differences are not given corresponding to 79°,
therefore required angle = 79°6′.
Math Class X
8
Question Bank
14. A boy standing on a vertical cliff in a jungle observes two rest
houses in line with him on opposite sides deep in the jungle below.
If their angles of depression are 19° and 26° and the distance
between them is 222 m, find the height of the cliff.
Solution. Let A be the top of the cliff and C and D be the two rest
houses.
Let AB = h m and BC = x m
Then,
BD = (222 – x) m
h
In ΔABC, tan 19° =
x
h
⇒
0.3443
=
x
h
= x × 0.3443
..(i)
⇒
h
In ΔABD, tan 26° =
222 – x
h
⇒
0.4877
=
222 – x
h
= (222 – x) × 0.4877
..(ii)
⇒
From (i) and (ii), we have
(222 – x) × 0.4877 = x × 0.3443
⇒ x (0.3443 + 0.4877) = 222 × 0.4877
222 × 0.4877
x =
⇒
0.832
222 × 0.4877
× 0.3443
From (i), we have, h =
0.832
⇒ log h = log 222 + log 0.4877 + log 0.3443 – log 0.832
–
–
–
= 2.3463 + 1.6881 + 1.5369 – 1.9201
= 2.3463 + (–1 – 1 + 1) + (0.6881 + 0.5369 – 0.9201)
= 2.3463 – 1 + 0.3049 = 2.3463 – 0.6951 = 1.6512
h = antilog 1.6512 = 44.79
⇒
Hence, height of the cliff = 44.80 m.
Math Class X
9
Question Bank
13. An aeroplane is flying horizontally 4000 m above the ground and is
going away from an observer on the level ground. At a certain
instant the observer finds that the angle of elevation of the plane is
45°. After 15 seconds, its elevation from the same point changes to
30°. Find the speed of the aeroplane in km/h.
Solution. Let A be the position
of the observer, B be the point
whose angle of elevation from
A is 45°. Let after 15 seconds
the position of the plane be C,
whose angle of elevation from A be 30°.
BD
In ΔABD, tan 45° =
AD
BD
⇒
1 =
AD
⇒
BD = AD
..(i)
CE
4000
[∴ AE = AD + DE]
=
In ΔACE, tan 30° =
AE
AD + DE
1
4000
=
[From (i)]
⇒
BD + DE
3
1
4000
=
⇒
4000 + DE
3
⇒
4000 3 = 4000 + DE
⇒
DE = 4000 3 – 1
(
)
⇒
DE = 4000 × 0.732 = 2928
⇒ Distance covered by the aeroplane in 15 seconds = 2928 m
2928
2928 18
×
⇒ Speed of the aeroplane =
m/s =
km/h
15
5
15
= 702.72 km/h.
Math Class X
10
Question Bank
14. At the foot of a mountain, the elevation of its summit is 45°. After
ascending 1000 m towards the mountain up a slope of 30°
inclination, the elevation is found to be 60°. Find the height of the
mountain.
Solution. Let AB be the mountain of height h m and C be its foot.
CD = 1000 m, ∠ACB = 45°, ∠DCB = 30° and ∠ADF = 60°.
h
In ΔACB, tan 45° =
CB
h
⇒
1 =
CB
⇒
h = CB
..(i)
DE
In ΔCDE, sin 30° =
1000
1
DE
⇒
=
2
1000
⇒
DE = 500
..(ii)
CE
In ΔCDE, cos 30° =
1000
CE
3
⇒
=
1000
2
⇒
CE = 500 3
..(iii)
Now, BE = BC – EC = h – 500 3
[From (i) and (iii)]
AF
In Δ ADF, tan 60° =
DF
h – BE
h – 500
⇒
3=
= [∵ DF = BF and BF = DE]
=
h – 500 3
h – 500 3
⇒
h – 500 = h 3 – 1500
⇒
h 3 –1
= 1000
(
⇒
Math Class X
h=
)
1000
= 1369.86
0.73
11
Question Bank
Hence, height of the mountain is 1369.86 m.
15. A man is standing on the deck of a ship which is 8 m above water
level. He observes the angle of elevation of the top of a hill as 60°
and the angle of depression of the base of the hill as 30°. Calculate
the distance of the hill from the ship and the height of the hill.
Solution. In the figure. A is the deck of the ship and CD is the hill.
Let BC = x m and DE = h m.
8
In ΔABC, tan 30° =
x
1
8
⇒
= ⇒ x = 8 3 m.
x
3
DE h
In ΔADE, tan 60° =
=
x
AE
h
⇒
3 =
[AE = BC = x]
x
⇒
h = 3x = 3 × 8 3 = 24 cm.
Distance of the hill from the ship = 8 3 m, and height of the hill
= (h + 8) m = (24 + 8) m = 32 m.
16. A ladder rests against a house on one side of a street. The angle of
elevation of the top of the ladder is 60°. The ladder is turned over to
rest against a house on the other side of the street and the elevation
now becomes 42° 50'. If the ladder is 40 m long, find the breadth of
the street.
Solution. In the figure, AB and CD are two houses. O is a point on
the street, at which one end of the ladder rests.
Let
OB = x m and OD = y m.
x
In ΔAOB, cos 60° =
40
1
x
⇒
=
⇒ x = 20 m.
2 40
Math Class X
12
Question Bank
y
40
y
⇒
0.7333 =
[From tables]
40
⇒ y = 0.7333 × 40 = 29.332
Hence, breadth of the street = (x + y) m
= (20 + 29.33) m = 49.33 m.
17. A vertical tower stands on horizontal plane and is surmounted by a
vertical flagstaff of height h m. At a point on the plane, the angle of
elevation of the bottom of the flagstaff is α and that of the top of the
h tan α
.
flagstaff is β. Prove that the height of the tower is
tanβ – tanα
Solution. Let AB be the tower, AC be the flagstaff of height h m
and D be the point of observation.
AB
In ΔABD, tan α =
BD
⇒
AB = BD tan α
..(i)
BC
In ΔCBD, tan β =
BD
AB + AC
⇒
tan β =
BD
⇒
AB + AC = BD tanβ
⇒
BD tan α + h = BD tanβ
⇒
h
= BD (tanβ – tanα)
h
⇒
BD =
tanβ – tanα
∴ Height of the tower = AB = BD tan α
[From (i)]
h tan α
=
tanβ – tanα
In ΔCOD, cos 42°50' =
Math Class X
13
Question Bank