ID : th-10-Trigonometry [1]
Grade 10
Trigonometry
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Answer t he quest ions
(1)
If 2sinθ – cosθ = 2, f ind the value of sinθ + 2cosθ.
(2)
Simplif y
(3)
Simplif y
(4) Simplif y
(5)
.
If sec θ = a/b and 0° > θ > 90°, f ind value of tan θ.
(6) From a tower on a straight road, the angles of depression of two cars at an instant are 45° and
60°. If the cars are 20 m apart, f ind the height of the tower.
Choose correct answer(s) f rom given choice
(7) T he value of sin x increases f aster than tan x as x increases (x < 90°).
(8)
a. T rue
b. False
c. Depends on value of θ
d. Depends on increment in θ
Simplif y
a. 1/secθ
b. 2 cotθ
c. 2 tanθ
d. 2 secθ
(9)
=?
a. 1 + cot θ
b. tanθ + cot θ
c. tanθ - cot θ
d. cotθ - tan θ
(10) If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is
decreasing.
a. T rue
b. False
c. Depends on the height of tower
d. Depends on value of angle
(11) If sin x + sin2x = 1, then f ind the value of expression (cos 2x + cos 4x).
a. 1
b. 0
c. 2
d. 0.5
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ID : th-10-Trigonometry [2]
(12)
Simplif y
a. 1 + tanθ - cotθ
b. sin2θ - cos 2θ
c. 0
d. 1 - sinθ cosθ
(13)
tan2θ
Simplif y 1 +
.
1 + secθ
a. tan θ
b. sec θ
c. cot θ
d. cosec θ
(14) Simplif y cos 3θ sin3(90°-θ) + sin3θ cos 3(90°-θ) + 3 cos 2θ sin2θ
a. tanθ + cotθ
b. sinθ cosθ
c. sinθ + cosθ
d. 1
Fill in t he blanks
(15)
sin257° + sin233°
+ sin233° + sin 57° cos 33° =
.
cos 257° + cos 233°
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ID : th-10-Trigonometry [3]
Answers
(1)
1
Step 1
It is given that
2sinθ – cosθ = 2
⇒ (2sinθ – cosθ)2 = 22 ....... [On squaring both sides]
Step 2
Now add (sinθ + 2cosθ)2 to both sides of equation
⇒ (2sinθ – cosθ)2 + (sinθ + 2cosθ)2 = 22 + (sinθ + 2cosθ)2
⇒ 4sin2θ + cos 2θ - 4sinθcosθ + sin2θ + 4cos 2θ + 4sinθcosθ = 4 + (sinθ + 2cosθ)2
⇒ 5sin2θ + 5cos 2θ + 4sinθcosθ - 4sinθcosθ = 4 + (sinθ + 2cosθ)2
⇒ 5(sin2θ + cos 2θ) = 4 + (sinθ + 2cosθ)2
⇒ 5 = 4 + (sinθ + 2cosθ)2
⇒ (sinθ + 2cosθ)2 = 5 - 4
⇒ (sinθ + 2cosθ)2 = 1
⇒ sinθ + 2cosθ = ± 1
Step 3
Since 2sinθ – cosθ=2, ignore the negative value.
T heref ore, sinθ + 2cosθ = 1
(2)
2 secθ
Step 1
=
Step 2
=
Step 3
=
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ID : th-10-Trigonometry [4]
(3)
sinθ + cosθ
Step 1
cosθ
We have been asked to simplif y the
sinθ
+
1 - tanθ
.
1 - cotθ
Step 2
Now,
cosθ
cosθ
sinθ
+
1 - tanθ
=
cos 2θ
sin2θ
sinθ
cosθ
+
1-
cosθ
sinθ
sinθ - cosθ
cos 2θ
-
sinθ - cosθ
=
1-
sin2θ
+
cosθ - sinθ
=
=
1 - cotθ
sinθ
sinθ - cosθ
sin2θ - cos 2θ
sinθ - cosθ
=
(sinθ - cosθ)(sinθ + cosθ)
sinθ - cosθ
= sinθ + cosθ
Step 3
T heref ore,
cosθ
+
1 - tanθ
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sinθ
is equal to the sinθ + cosθ.
1 - cotθ
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ID : th-10-Trigonometry [5]
(4) cosec θ + cot θ
Step 1
We know that 1 - cos 2θ = sin2θ. Using this identity we can simplif y denominator, if we
multiply it by 1 + cosθ
Step 2
Now on multiplying numerator and denominator by 1 + cosθ,
√(
1 + cosθ
) = √(
1 - cosθ
= √(
1 + cosθ
×
1 - cosθ
(1 + cosθ)2
1 + cosθ
)
1 + cosθ
)
1 - cos 2θ
= √(
(1 + cosθ)2
)
sin2θ
=
1 + cosθ
sinθ
=
1
+
sinθ
cosθ
sinθ
= cosecθ + cotθ
(5)
Step 1
We know that,
tan θ = √(sec2θ - 1)
Step 2
Now replace value of sec(θ) in above equation
⇒ tan(θ) =
Step 3
Simplif y RHS of above equation
⇒ tan(θ) =
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ID : th-10-Trigonometry [6]
(6) 10(3 + √3 ) m
Step 1
Following f igure shows the tower and the cars,
Let's assume the height of the tower(AB) be 'y' and the distance between the tower and the
f arther car i.e BC be 'x'.
According to the question, the distance between the cars is 20 m,
theref ore, BD = (x - 20) m
Step 2
In right angled triangle ABD,
AB
= tan60°
BD
y
⇒
= √3
x - 20
⇒ y = √3(x - 20) -----(1)
Step 3
In right angled triangle ABC,
AB
= tan45°
BC
⇒
y
=1
x
⇒ y = x -----(2)
Step 4
By putting the value of 'x' f rom equation (2) to equation (1), we get:
y = 10(3 + √3 )
Step 5
T hus, the height of the tower is 10(3 + √3 ) m.
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ID : th-10-Trigonometry [7]
(7) b. False
Step 1
Let's observe this right angle triangle,
sin x =
BC
AC
tan x =
BC
AB
Step 2
Now let's increase the angle x such that height BC remains unchanged,
Step 3
By comparing above two pictures, we can notice that decrease in AB is larger than decrease
in AC
Step 4
Now numerator of both the f ractions are unchanged, but denominator of tan x is
decreasing f aster than denominator of sin x.
T heref ore tan x increases f aster than sin x, as x increases
Step 5
Hence given statement is False
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ID : th-10-Trigonometry [8]
(8)
b. 2 cotθ
Step 1
=
Step 2
=
Step 3
=
(9) b. tanθ + cot θ
Step 1
Using identities sec2θ = 1 + tan2θ and cosec2θ = 1 + cot 2θ
S=
⇒S=
Step 2
Using the relation tanθcotθ = 1, we can re-write above expression as f ollowing
⇒S=
Step 3
Using (a + b)2 = a2 + b2 + 2ab
⇒S=
⇒ S = tanθ + cotθ
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ID : th-10-Trigonometry [9]
(10) a. T rue
Step 1
Let's assume, BD is the lengths of the shadow of the tower AB and the length of the
shadow of the tower is increasing f rom BD to BC, as shown in the f ollowing f igure,
Step 2
It can be seen that angle ∠x > ∠y, theref ore as the shadow of the tower is increasing, the
angle of elevation of the sun is decreasing
Step 3
T heref ore, given statement is T rue.
(11) a. 1
Step 1
It is given that
sin x + sin2x = 1 ............. Eq. (1)
Step 2
Above equation can be re-written as f ollowing
⇒ 1 - sin2x = sin x
Step 3
Since 1 - sin2θ = cos 2θ, we can re-write equation as f ollowing,
⇒ cos 2x = sin x ............. Eq. (2)
Step 4
Now replace value of cos 2x = sin x, in required expression
cos 2x + cos 4x = sin x + sin2x
⇒ cos 2x + cos 4x = 1 .................... Using Eq. (1)
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ID : th-10-Trigonometry [10]
(12) c. 0
Step 1
On adding two f ractions
=
Step 2
=
Step 3
=
Step 4
=
(13) b. sec θ
Step 1
Let, S = 1 +
tan2θ
1 + secθ
Step 2
Using identity tan2θ = sec2θ - 1,
⇒S= 1+
sec2θ - 1
1 + secθ
Step 3
Using a2 - b2 = (a+b)(a-b),
⇒S= 1+
(secθ + 1)(secθ - 1)
1 + secθ
⇒ S = 1 + (secθ - 1)
⇒ S = secθ
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ID : th-10-Trigonometry [11]
(14) d. 1
Step 1
Since sin(90°-θ ) = cos θ and cos(90°-θ ) = sin θ , expression can be rewritten as f ollowing
= (cos 2θ)3 + (sin2θ)3 + 3 cos 2θ sin2θ
Step 2
Since x3 + y3 = ( x + y ) ( x2 + y2 - xy)
= (cos 2θ + sin2θ ) [ (cos 2θ)2 + (sin2θ)2 - cos 2θ sin2θ ] + 3 cos 2θ sin2θ
Step 3
Since cos 2θ + sin2θ =1 and x2 + y2 = (x+y)2 - 2 xy)
= (cos 2θ + sin2θ ) [ (cos 2θ + sin2θ)2 - 2 cos 2θ sin2θ - cos 2θ sin2θ] + 3 cos 2θ sin2θ
= [ 1 - 3 cos 2θ sin2θ ] + 3 cos 2θ sin2θ
=1
(15)
2
Step 1
Lets,
sin257° + sin233°
S=
+ sin233° + sin 57° cos 33°
cos 257° + cos 233°
Step 2
Using identities sin(θ) = cos(90° - θ) and cos(θ) = sin(90° - θ), we can re-write expression
as f ollowing
sin257° + cos 2(90°-33°)
⇒S=
+ sin233° + cos (90°-57°) cos 33°
cos 257° + cos 2(90°-33°)
sin257° + cos 257°
⇒S=
+ sin233° + cos 33° cos 33°
cos 257° + cos 257°
Step 3
Using identity sin2θ + cos 2θ = 1
⇒ S = 1/1 + sin233° + cos 233°
⇒S= 1+ 1
⇒S= 2
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