ID : gb-10-Trigonometry [1]
Grade 10
Trigonometry
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Answer t he quest ions
(1)
(2)
(3)
Simplif y
.
Simplif y
An equilateral triangle is inscribed in a circle of radius 8 cm. Find the length of triangle's side.
(4) From a tower on a straight road, the angles of depression of two cars at an instant are 30° and
45°. If the cars are 10 m apart, f ind the height of the tower.
Choose correct answer(s) f rom given choice
(5)
(cosθ + secθ)2 + (sinθ + cosecθ)2 = ?
a. 7 + tan2θ - cot 2θ
b. 7 + tan2θ + cot 2θ
c. 7 + cot 2θ - tan2θ
d. 6 + cot 2θ - tan2θ
(6) (1 + cotθ - cosecθ) (1 + tanθ + secθ) = ?
a. 0
b. -2
c. 1
d. 2
(7) If cos 5A = sin A and 5A < 90°, f ind value of tan 4A.
a.
b.
c.
d.
(8)
Simplif y
sin4x - cos 4x + 1
sin2x
a. 1
b. -1
c. 2
d. 3
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ID : gb-10-Trigonometry [2]
(9)
=?
a. 1 - sinθ cosθ
b. tanθ + cotθ
c. 1
d. 2
(10) cosec4θ - cosec2θ
a. cot 2θ + cot 4θ
b. tan2θ + cot 4θ
c. cot 2θ - cot 4θ
d. cot 2θ + tan4θ
(11) Simplif y
a. 0
b. 2 secθ
c. 2 tanθ
d. 1/secθ
(12) AD =
cm, DC =
cm and DB =
cm. Find the value of angle ∠ACB.
a. 110°
b. 100°
c. 95°
d. 105°
(13) If cos x =
and sin y =
(0° < x < 90° and 0° < y < 90°), f ind value of tan 4(y-x).
a.
b.
c.
d.
(14) Simplif y
a. 1 + sinθ cosθ
b. tanθ + cotθ
c. cos 2θ - sin2θ
d. 1 - sinθ cosθ
(15) If cos(X+Y) = 0, sin(X-Y) = ?
a. cos Y
b. sin X
c. cos 2Y
d. sin 2X
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ID : gb-10-Trigonometry [3]
Answers
(1)
2cosec θ
S=
⇒S=
⇒S=
⇒S=
⇒ S = 2cosec θ
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ID : gb-10-Trigonometry [4]
(2)
sin2x cos 2x
Step 1
(1 + cotx + tanx)(sinx - cosx)
We have been asked to simplif y the
sec3x - cosec3x
.
Step 2
(1 +
(1 + cotx + tanx)(sinx - cosx)
=
sec3x - cosec3x
cosx
+
sinx
sinx
)(sinx - cosx)
cosx
)
(secx - cosecx)(sec2x + secx × cosecx + cosec2x)
[Since, a3 - b3 = (a - b)(a2 + ab + b2]
(
sinx cosx + cos 2x + sin2x
)(sinx - cosx)
sinx cosx
=
(
1
1
-
cosx
)(
sinx
1
cosecx =
1
cos 2x
[Since, secx =
1
+
sinx cosx
+
1
cos 2x
1
and
cosx
)
]
sinx
(
1 + sinx cosx
)(sinx - cosx)
sinx cosx
=
(sinx - cosx)(sin2x + sinx cosx + cos 2x)
(sinx cosx)(sin2x cos 2x)
=
(1 + sinx cosx)(sinx - cosx)(sin3x cos 3x)
(sinx cosx)(sinx - cosx)(1 + sinx cosx)
= sin2x cos 2x
Step 3
T heref ore,
(1 + cotx + tanx)(sinx - cosx)
sec3x - cosec3x
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is equal to the sin 2x cos2x.
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ID : gb-10-Trigonometry [5]
(3)
8√3 cm
Step 1
T riangle ABC is inscribed in a circle. Lets O be the center of circle.
Step 2
Now connect all vertices of the triangle to O, and draw a perpendicular f rom O to side BC
of triangle at point D.
Step 3
OB and OC are bisectors of ∠B and ∠C respectively, hence
∠OBD = 30°
Step 4
Since ∠ ODB is right angle
BD/OB = cos 30° = √3/2
Step 5
BD = OB × (√3/2) = 8 × (√3/2) = 4√3
Step 6
BC = 2 BD = 2(4√3) = 8√3 cm
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ID : gb-10-Trigonometry [6]
(4) 5(1 + √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 10 m,
theref ore, BD = (x - 10) m
Step 2
In right angled triangle ABD,
AB
= tan45°
BD
y
⇒
=1
x - 10
⇒ y = x - 10 -----(1)
Step 3
In right angled triangle ABC,
AB
= tan30°
BC
⇒
y
=
x
1
√3
⇒ √3(y) = x -----(2)
Step 4
By putting the value of 'x' f rom equation (2) to equation (1), we get:
y = 5(1 + √3)
Step 5
T hus, the height of the tower is 5(1 + √3) m.
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ID : gb-10-Trigonometry [7]
(5)
b. 7 + tan2θ + cot 2θ
Step 1
Let's expand using the algebraic identity (a + b)2 = a2 + b2 + 2ab
(cosθ + secθ)2 + (sinθ + cosecθ)2,
= (cosθ)2 + (secθ)2 + 2 cosθ secθ + (sinθ)2 + (cosecθ)2 + 2 sinθ cosecθ
= sin2θ + cos 2θ + sec2θ + cosec2θ + 2 + 2
Step 2
Now using trigonometrical identities sin2θ + cos 2θ =1
= 5 + sec2θ + cosec2θ
Step 3
Now let's use trigonometrical identities sec2θ = 1 + tan2θ and cosec2θ = 1 + tan2θ
= 5 + 1 + tan2θ + 1 + cot 2θ
= 7 + tan2θ + 1 + tan2θ
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ID : gb-10-Trigonometry [8]
(6) d. 2
Step 1
We need to f ind f ollowing product
S = (1 + cotθ - cosecθ) (1 + tanθ + secθ)
Step 2
On multiplying each terms
⇒ S = 1 (1 + tanθ + secθ) + cotθ (1 + tanθ + secθ) - cosecθ (1 + tanθ + secθ)
⇒ S = (1 + tanθ + secθ) + (cotθ + cotθ tanθ + cotθ secθ ) - (cosecθ + cosecθ tanθ +
cosecθ secθ)
Step 3
Using identities cotθ tanθ = 1, cotθ secθ = cosecθ and cosecθ tanθ = secθ
⇒ S = (1 + tanθ + secθ) + (cotθ + 1 + cosecθ ) - (cosecθ + secθ + cosecθ secθ)
Step 4
Now positive cosecθ and secθ will cancel each other
⇒ S = (1 + tanθ + secθ) + (cotθ + 1 + cosecθ ) - (cosecθ + secθ + cosecθ secθ)
⇒ S = 2 + tanθ + cotθ - cosecθ secθ
Step 5
Using identities tanθ = sinθ/cosθ, and cotθ = cosθ/sinθ
⇒S= 2+
sinθ
+
cosθ
⇒S= 2+
cosθ
- cosecθ secθ
sinθ
sin2θ + cos 2θ
- cosecθ secθ
sinθcosθ
⇒S= 2+
1
- cosecθ secθ
sinθcosθ
⇒ S = 2 + cosecθ secθ - cosecθ secθ
⇒S= 2
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ID : gb-10-Trigonometry [9]
(7) c.
Step 1
It is given that,
cos 5A = sin A .......... Eq.(1)
Step 2
We know that sin(90°-θ ) = cos(θ )
T heref ore (1) can be re-written as f ollowing,
sin(90°-5A) = sin A .......... Eq.(2)
Step 3
On comparing two sides of (2), we get
⇒ 90°-5A = A
⇒ 90° = (1 + 5) A
⇒ A = 90°/6
⇒ A = 15°
Step 4
Now replace the value of A in tan 4A,
⇒ tan 4A = tan(4 × 15)
⇒ tan 4A = tan(60°)
⇒ tan 4A =
(8)
c. 2
Step 1
Lets f irst simplif y sin4x - cos 4x + 1
S = sin4x - cos 4x + 1
⇒ S = sin4x - (cos 2x)2 + 1
⇒ S = sin4x - (1 - sin2x)2 + 1 ........................ [Using identity cos 2θ = 1 sin2θ]
⇒ S = sin4x - (1 + sin4x - 2 sin2x) + 1 ........... [Using identity (a - b)2 = a2 + b2 - 2ab]
⇒ S = sin4x - 1 - sin4x + 2 sin2x + 1
⇒ S = sin4x - 1 - sin4x + 2 sin2x) + 1
⇒ S = 2 sin2x
Step 2
T heref ore
sin4x - cos 4x + 1
=
sin2x
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2 sin2x
=2
sin2x
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ID : gb-10-Trigonometry [10]
(9) d. 2
Step 1
Since sin(90°-θ ) = cos θ and cos(90°-θ ) = sin θ , expression can be rewritten as f ollowing
=
cos 240° + sin240°
+
cosθ sinθ
sin240° + cos 240°
tanθ
+
sinθ cosθ
cotθ
Step 2
= 1 + cos 2θ + sin2θ
Step 3
=1+1=2
(10) a. cot 2θ + cot 4θ
Step 1
Let,
S = cosec4θ - cosec2θ
Step 2
Lets take cosec2θ common f rom both terms,
⇒ S = cosec2θ (cosec2θ -1)
Step 3
Using trigonometrical identity cosec2θ = 1 + cot 2θ
⇒ S = (1 + cot 2θ) cot 2θ
⇒ S = cot 2θ + cot 4θ
(11) b. 2 secθ
Step 1
=
Step 2
=
Step 3
=
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ID : gb-10-Trigonometry [11]
(12) d. 105°
Step 1
tan A = DC/AD
Step 2
tan A = 1/√3
Step 3
A = 30°
Step 4
Similarly tan B = DC/DB
Step 5
tan B = 1
Step 6
B = 45°
Step 7
∠ACB = 180° - (∠CAB + ∠ABC) = 180° - (30° + 45°) = 105°
(13) c.
Step 1
We know that cos(30°) =
, and sin(45°) =
Step 2
On comparing we get,
x = 30°
y = 45°
Step 3
On substituting values of x and y
tan 4(y-x) = tan (4(45° - 30°))
⇒ tan 4(y-x) = tan (60°)
⇒ tan 4(y-x) =
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ID : gb-10-Trigonometry [12]
(14) a. 1 + sinθ cosθ
Step 1
Multiply numerator and denominator of f irst term by cosθ
=
Step 2
Since x3 - y3 = ( x + y ) ( x2 + x2 + xy)
=
Step 3
Now (cosθ - sinθ) cancels out and since sin2θ + cos 2θ = 1
=
(15) c. cos 2Y
Step 1
We are given that cos(X + Y) = 1, but we already know that
cos 90° = 0
Step 2
T his means, X + Y = 90°
=> X = 90° - Y ........(1)
Step 3
Now replace value of X f rom eq. (1) in sin(X-Y) :
sin(X-Y)
= sin(90° - Y - Y)
= sin(90° - 2Y)
= cos(2Y) ......... [ Here we have applied the identity sin(90° - θ ) = cos(θ ) ]
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