1. If sin θ + cosec θ = 2, then [latex]{sin}^{2}θ + {cosec}^{2}θ [/latex]is equal to:

A. 1 B. 4 C. 2 D. None of these

Answer - Option C
Explanation -
Given that, sinθ + cosecθ = 2
On squaring both sides, we get
[latex]{sin}^{2}θ + {cosec}^{2}θ [/latex] + 2 = 4
[latex]{sin}^{2}θ + {cosec}^{2}θ [/latex] = 2
2. The value of [latex]\frac{{cot}^{2}θ + 1}{{cot}^{2}θ - 1}[/latex] is equal to:

A. sin 2θ B. cos 2θ C. cosec 2θ D. sec 2θ

Answer - Option D
Explanation -
[latex]\frac{{cot}^{2}θ + 1}{{cot}^{2}θ - 1}[/latex] = [latex]\frac{+ 1{tan}^{2}θ}{1 - + 1{tan}^{2}θ}[/latex]
[latex]\frac {1}{{cos}^{2}θ + {sin}^{2}θ} [/latex] = [latex]\frac{1}{cos2θ}[/latex] = sec2θ
3. If tan A = 2 tan B + cot B, then 2 tan (A – B) is equal to:

A. tan B B. 2 tan B C. cot B D. 2 cot B

Answer - Option C
Explanation -
Given that
tan A = 2 tan B + cot B ...(i)
Now, 2 tan (A – B) = 2([latex]\frac{tan A - tan B}{1 + tanA tan B}[/latex])
[latex]2 \frac{(2tan B + cot B - tan B)}{1 + (2tan B cot B) tan B}[/latex]
[latex]2 \frac{tan B cot B}{2(1 + {tan}^{2} B)}[/latex] = [latex]\frac{cotB({tan}^{2} B + 1)}{1 + {tan}^{2} B}[/latex] = cotB
4. If tan A – tan B = x and cot B – cot A = y, then cot (A – B) is equal to:

A. [latex]\frac{1}{x} + y[/latex] B. [latex]\frac{1}{xy}[/latex] C. [latex]\frac{1}{x} - \frac{1}{y}[/latex] D. [latex]\frac{1}{x} + \frac{1}{y}[/latex]

Answer - Option D
Explanation -
Given that
tan A – tan B = x ...(i)
and cot B – cot A = y ...(ii)
Now, cot (A – B) = [latex]\frac{1}{tan (A - B)}[/latex]
= [latex]\frac{1 + tan A tan B}{tan A - tan B}[/latex]
= [latex]\frac{1}{tan A - tan B)} + \frac{tan A tan B}{tan A - tan B}[/latex]
= [latex]\frac{1}{x} + \frac{1}{y}[/latex] [from (i) and (ii)]
5. If sin θ = [latex]\frac{4}{5}[/latex] and θ lies in the third quadrant, then cos [latex]\frac{θ}{2}[/latex] is equal to:

A. [latex]\frac{1}{\sqrt{5}}[/latex] B. [latex] - \frac{1}{\sqrt{5}}[/latex] C. [latex]\sqrt{\frac{2}{5}}[/latex] D. -[latex]\sqrt{\frac{2}{5}}[/latex]

Answer - Option B
Explanation -
Given that,
sinθ = [latex] - \frac{4}{5}[/latex] = and θ lies in the IIIrd quadrent
cosθ = [latex]\sqrt{1 - \frac{16}{25}}[/latex]
Now, [latex] cos \frac{θ}{2}[/latex] = [latex]\sqrt{ \frac{1 + cosθ}{2}}[/latex] = [latex]\sqrt{ \frac{1 - \frac{3}{5} }{2}}[/latex] = ± [latex]\sqrt{\frac{1}{5}}[/latex]
But we take [latex] cos \frac{θ}{2}[/latex] = -[latex]\sqrt{\frac{1}{5}}[/latex] Since, if θ lies in IIIrd
quadrant, then [latex]\frac{θ}{2}[/latex] will be in IInd quadrant.
Hence, [latex] cos \frac{θ}{2}[/latex] = -[latex]\sqrt{\frac{1}{5}}[/latex]
6. The value of cos 1º cos 2º cos 3º... cos 100º is equal to

A. 0 B. -1 C. 0 D. None of these

Answer - Option C
Explanation -
cos 1º cos 2º cos 3º ... cos 90º... cos 100º
= cos 1º cos 2º cos 3º ... 0 ... cos 100º = 0
[i.e, cos 90º = 0]
7. The value of sin 12º sin 48º sin 54º is equal to:

A. [latex]\frac{1}{16}[/latex] B. [latex]\frac{1}{32}[/latex] C. [latex]\frac{1}{8}[/latex] D. [latex]\frac{1}{4}[/latex]

Answer - Option C
Explanation -
Now, sin 12º sin 48º sin 54º
[latex]\frac{1}{2}[/latex](cos36º - cos60º )cos36º
[latex]\frac{1}{2}(\frac{(\sqrt{5} + 1}{4}\frac{1}{2})(\frac{(\sqrt{5} + 1}{4})[/latex] = [latex]\frac{5 -1}{32} = \frac{4}{32} = \frac{1}{8}[/latex]
8. The value of 2 cos x – cos 3x – cos 5x is equal to:

A. [latex]16{cos}^{3}x {sin}^{2}x[/latex] B. [latex]16{sin}^{3}x {cos}^{2}x[/latex] C. [latex]4{cos}^{3}x {sin}^{2}x[/latex] D. [latex]4{sin}^{3}x {cos}^{2}x[/latex]

Answer - Option A
Explanation -
2 cos x – cos 3x – cos 5x = 2 cos x – 2 cos x cos 4x
= 2 cos x (1 – cos 4x) = 2 cos x 2 [latex]{sin}^{2}2x[/latex]
4 cos x[latex]{ (2 sin x cos x)}^{2}[/latex] = [latex]16 {sin}^{2}x{cos}^{3}x[/latex]
9. If cos θ = [latex]\frac{1}{2}(x + \frac{1}{x})[/latex] then [latex]\frac{1}{2}({x}^{2} + \frac{1}{{x}^{2}})[/latex] is equal to:

A. sin 2θ B. cos 2θ C. tan 2θ D. sec 2θ

Answer - Option D
Explanation -
Given that cos θ = [latex]\frac{1}{2}(x + \frac{1}{x})[/latex] = [latex]x + \frac{1}{x}[/latex] = 2 cosθ
We know that [latex]{x}^{2} + \frac {1}{{x}^{2}}[/latex] = [latex]{x}^{2} + \frac {1}{{x}^{2}}[/latex]
= [latex]{2cos θ}^{2} -2 = {4 cos θ}^{2}[/latex] -2
= 2cos 2θ [from (i)]
i.e, [latex]\frac{1}{2}({x}^{2} + \frac {1}{{x}^{2}})[/latex] =[latex] \frac{1}{2} * {x}^{2} 2cos 2θ[/latex]
10. The value of x for the maximum value of [latex]\sqrt{3}[/latex]cosx + sin x is:

A. 30º B. 45º C. 60º D. 90º

Answer - Option A
Explanation -
Let f(x) = [latex]\sqrt{3}[/latex]cosx + sin x
f(x) = [latex]2(\frac{\sqrt{3}}{2}cosx + \frac{1}{2}sinx) [/latex] = 2sin ([latex] x + \frac{\pi}{3}[/latex])
Since, -1 ≤ sin (x + [latex]\frac{\pi}{3}) ≤ 1[/latex]
Hence, f(x) is maximum if x + [latex]\frac{\pi}{3}[/latex] = [latex]\frac{\pi}{2}[/latex]
x = [latex]\frac{\pi}{6}[/latex] = 30º
11. The equation [latex]{(a + b)}^{2}[/latex] = 4ab [latex]{sin}^{2}θ [/latex] is possible only when

A. 2a = b B. a = b C. a = 2b D. None of these

Answer - Option A
Explanation -
We have [latex]{(a + b)}^{2}[/latex] = 4ab [latex]{sin}^{2}θ [/latex]
[latex]{sin}^{2}θ [/latex] = [latex]\frac{{(a + b)}^{2}}{4ab}[/latex]
Since [latex]{sin}^{2}θ [/latex] ≤ 1
= [latex]\frac{{(a + b)}^{2}}{4ab}[/latex] ≤ 1
= [latex]{(a + b)}^{2} - 4ab = {(a + b)}^{2} [/latex] ≤ 1
= [latex]{(a - b)}^{2} [/latex] ≤ 0
a = b
12. The value of the expression 1 - [latex]\frac{{sin}^{2}y}{1 + cos y} + \frac{1 + cos y}{sin y} - \frac {sin}{1 - cos y}is equal to:[/latex]

A. 0 B. 1 C. sin y D. cos y

Answer - Option D
Explanation -
The given expression can be written as
= [latex]\frac { 1 + cos y - {sin} ^ {2}y }{1 + cos y}[/latex] + [latex] \frac { 1 + {cos}^{2} y - {sin}^{2} y }{sin y (1 + cos y)} [/latex]
= [latex]\frac{cos y (1 + cos y)}{1 + cos y} + 0[/latex] = cos y
13. The circular wire of diameter 10 cm is cut and placed along the circumference of a circle of diameter 1m. The angle subtended by the write at the center of the circle is equal to:

A. [latex]\frac{\pi}{4}rad[/latex] B. [latex]\frac{\pi}{3}rad[/latex] C. [latex]\frac{\pi}{5}rad[/latex] D. [latex]\frac{\pi}{10}rad[/latex]

Answer - Option C
Explanation -
Given that, diameter of circular wire = 10 cm,
Length of wire = 10[latex]\ pi[/latex]
Hence, required angle = [latex]\frac{length of arc}{radius of big circle}[/latex]
[latex]\frac{10 \pi}{50}[/latex] = [latex]\frac{\pi}{5}[/latex]
14. The greatest and least value of sin x cos x are:

A. 1, -1 B. [latex]\frac{1}{2}[/latex], - [latex]\frac{1}{2}[/latex] C. [latex]\frac{1}{4}[/latex], - [latex]\frac{1}{4}[/latex] D. 2, -2

Answer - Option B
Explanation -
f(x) = sin x cos x = [latex]\frac{1}{2} sin 2x[/latex]
We know – 1 ≤ sin 2x ≤ 1
- [latex]\frac{1}{2}[/latex] ≤ [latex]\frac{1}{2} sin 2x[/latex] ≤ [latex]\frac{1}{2}[/latex]
Thus, the greatest and least value of f(x) are [latex]\frac{1}{2}[/latex] and -[latex]\frac{1}{2}[/latex] respectively
15. If A = [latex]{sin}^{2}θ + {cosec}^{4}θ [/latex], then for all real values of θ

A. 1 ≤ A ≤ 2 B. [latex]\frac{3}{4}[/latex] ≤ A ≤ 1 C. [latex]\frac{13}{16}[/latex] ≤ A ≤ 1 D. [latex]\frac{3}{4}[/latex] ≤ A ≤ [latex]\frac{13}{16}[/latex]

Answer - Option B
Explanation -
We have, A = [latex]{sin}^{2}θ + {cos}^{4}θ [/latex]
= [latex]{sin}^{2}θ + {cos}^{2}θ {cos}^{2}θ [/latex] ≤ [latex]{sin}^{2}θ + {cos}^{2}θ [/latex] (Since, [latex]{cos}^{2}θ[/latex] ≤ 1)
[latex]{sin}^{2}θ + {cos}^{4}θ [/latex] ≤ 1
A ≤ 1
Again, [latex]{sin}^{2}θ + {cos}^{4}θ [/latex] = 1 - [latex]{cos}^{2}θ + {cos}^{4}θ [/latex]
= [latex]{cos}^{4}θ - {cos}^{2}θ + 1 [/latex]
= [latex]{{cos}^{2}θ - \frac{1}{2}}^{2}[/latex] + [latex]\frac{3}{4}[/latex] ≥ [latex]\frac{3}{4}[/latex]
Hence [latex]\frac{3}{4}[/latex]≤ A ≤ 1

1. The value of [latex] {sin}^{2} 30º + {sin}^{2} 60º[/latex] is

A. 1 B. [latex]\frac{3}{2}[/latex] C. 2 D. [latex]\frac{3}{4}[/latex]

Answer - Option A
Explanation -
[latex] {sin}^{2} 30º + {sin}^{2} 60º[/latex] = [latex] {sin}^{2} 30º + {cos}^{2} 30º[/latex] = 1
2. tan 90° is undefined. As θ is increased from 89° towards 90°, value of tan θ tends to

A. 0 B. +oo C. 1 D. undefined

Answer - Option B
Explanation -
Value of tan q will ten A to + ∞ as tan 90° is + ∞
3. The angle of elevation of a ladder leaning against a wall is 60° i.e. ladder makes an angle of 60° with the ground. The foot of the ladder is 4.6 metres away from the wall. What is the length of this ladder?

A. 9.2 m B. 2.3 m C. 6.9 m D. 7.8 m

Answer - Option A
Explanation -
[latex]\frac{4.2}{L}[/latex] = COS 60º
L = 4.6 × 2 = 9.2 m
4. A kite is flying at a height of 75 m from the level ground, attached to a string inclined at 60° to the horizontal. The length of the string is

A. 50 [latex]\sqrt{2}[/latex] m B. 50 [latex]\sqrt{3}[/latex] m C. [latex]\frac{50}{\sqrt{2}}[/latex] D. [latex]\frac{50}{\sqrt{3}}[/latex]

Answer - Option B
Explanation -
[latex]\frac{75}{L}[/latex] = cos60º
L = [latex]\frac{75 * 2}{\sqrt{3}}[/latex] = 50 [latex]\sqrt{3}[/latex] m
5. A tower stands vertically on the ground. From a point on the ground which is 30 m away from the foot of the lower, the angle of elevation of the top of the tower is 60°. The height of the tower is:

A. 30 [latex]\sqrt{3}[/latex] m B. 10 [latex]\sqrt{3}[/latex] m C. 15m D. 6m

Answer - Option A
Explanation -
Figure
tan 60° = [latex]\frac{Height}{30}[/latex]
Height of the tower = 30 [latex]\sqrt{3}[/latex] m
6. From a point on the bridge across a river the angles of depression of the banks on the opposite side of the river are 30° and 45° respectively. If the bridge is at a height of 4 m from the bank, the width of the river is

A. 4 ([latex]\sqrt{3} - 1[/latex]) m B. 2 ([latex] 2\sqrt{3} - 1[/latex]) m C. 4 ([latex]\sqrt{3} + 1[/latex]) m D. 2 ([latex] 2\sqrt{3} + 1[/latex]) m

Answer - Option C
Explanation -
Figure
In triangle ABD, tan 30° = [latex]\frac{DB}{AB}[/latex]
AB = 4 [latex]\sqrt{3}[/latex] m
Similarly, in triangle CBD, tan 45° = [latex]\frac{DB}{BC}[/latex]
BC = DB = 4 meter
Hence, required distance AC = 4 [latex]\sqrt{3}[/latex] + 4
= 4 ([latex]\sqrt{3} + 1[/latex]) m
7. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, the height of the wall is

A. 7.5 m B. 5 [latex] \sqrt{3}[/latex]) m C. 15 ([latex]\frac {\sqrt{3}}{2}[/latex]) m D. 10 ([latex] \sqrt{3}[/latex]) m

Answer - Option A
Explanation -
cos60 = [latex]\frac{h}{15}[/latex]
h = 15 * [latex]\frac{1}{2}[/latex] = 7.5 m
8. From a point on the bridge across a river the angles of depression of the banks on the opposite side of the river are 30° and 45° respectively. If the bridge is at a height of 3m from the bank, the width of the river is

A. 2 ([latex] \sqrt{3} + 1[/latex]) m B. 3 ([latex] \sqrt{3} + 1[/latex]) m C. 4([latex] \sqrt{3} + 1[/latex]) m D. 2 ([latex] 2\sqrt{3} + 1[/latex]) m

Answer - Option B
Explanation -
According to the figure.
In △ACD, [latex]\frac{AC}{CD}[/latex] = tan 45º
AC = CD
or CD = 3m
Again in △ACB, [latex]\frac{AC}{BC}[/latex] = tan 30º
BC = 3([latex] \sqrt{3}[/latex])
Hence, BD = 3 ([latex] \sqrt{3}[/latex]) + 3([latex] \sqrt{3} + 1[/latex]) m
9. A ladder just reaches the top of a wall. The foot of the ladder is 8 m away from the foot of the wall. The ladder makes an angle of 60° with the ground. The length of the ladder is

A. 4 m B. 16 m C. 16 [latex]\frac{\sqrt{3}}{3} [/latex] D. 16 [latex] \sqrt{3} [/latex]

Answer - Option B
Explanation -
cos 60° = [latex]\frac{8}{1}[/latex]
[latex]\frac{1}{2}[/latex] = [latex]\frac{8}{2}[/latex]
1 = 16 m
10. From a point on the bridge across a river the angle of depressions of the bank on the opposite side of the river are 60° and 45° respectively. If the bridge is at a height of 3 m from the bank, the width of the river is

A. 2 ([latex] \sqrt{3} + 1[/latex]) m B. 3 ([latex] \sqrt{3} + 1[/latex]) m C. 3 - ([latex] \sqrt{3} [/latex]) m D. 3 + ([latex] \sqrt{3} [/latex]) m

Answer - Option D
Explanation -
From the figure, tan 60° = [latex]\frac{3}{X}[/latex] = [latex]\sqrt{3}[/latex]
x = [latex]\sqrt{3}[/latex]
tan 45° = [latex]\frac{3}{y}[/latex] = 1
y = 3
i.e, Width of river = n + y = 3 + [latex]\sqrt{3}[/latex] m
11. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, the distance of the foot of the ladder from the wall is

A. 7.5 m B. [latex]5 \sqrt{3}[/latex] m C. [latex]10 \sqrt{3}[/latex]m D. [latex]\frac{15 \sqrt{3}}{2}[/latex] m

Answer - Option D
Explanation -
Let distance of foot of ladder from wall = x
sin 60° = [latex]\frac{x}{15}[/latex] = [latex]\frac{\sqrt{3}}{2}[/latex]
x = [latex]15 \frac{\sqrt{3}}{2}[/latex]
12. From a point on the bridge across a river the angles of depression of the bank on the opposite side of the river are 60° and 45° respectively. If the bridge is at a height of 4 m from the bank, the width of the river is

A. [latex] 4 \frac {3 + \sqrt{3}}{3}[/latex] m B. [latex] 4 (3 + \sqrt{3})[/latex] m C. [latex] 2 \frac{1 + \sqrt{3}}{3}[/latex] m D. [latex] 2 (1 + 2 \sqrt{3})[/latex] m

Answer - Option A
Explanation -
From the figure,
tan60º = [latex]\frac{4}{x}[/latex] = [latex]\sqrt{3}[/latex]
x = [latex]\frac{4}{\sqrt{3}}[/latex]
tan45º = [latex]\frac{4}{y}[/latex] = 1
y = 4
i.e, width of river = x + y
= [latex]\frac{4}{\sqrt{3}}[/latex] + 4
= [latex] 4 \frac {3 + \sqrt{3}}{3}[/latex] m
13. The angle of elevation of the top of a tower from a point on the ground Inch is 30 m away from the foot of the tower is 45°. The height of the tow t is

A. 15 m B. 10 [latex]\sqrt{3}[/latex] m C. 30 m D. 30 [latex]\sqrt{3}[/latex] m

Answer - Option C
Explanation -
tan 45° = [latex]\frac{h}{30}[/latex] = 1
h = 30 m
14. From a point on the bridge across a river the angle of depressions of the bank on the opposite side of the river are 30° and 60° respectively. If the bridge is at a height of 3 m from the bank, the width of the river is

A. 4 [latex]\sqrt{3}[/latex] m B. 2 [latex](\sqrt{3} + 1)[/latex] m C. 2 [latex](\sqrt{3} + 3)[/latex] m D. 2 [latex]\sqrt{3} [/latex] m

Answer - Option A
Explanation -
tan 60º = [latex]\frac{3}{x}[/latex] = [latex]\sqrt{3}[/latex]
x = [latex]\sqrt{3}[/latex]
tan 45º = [latex]\frac{3}{y}[/latex] = 1
y = 3
i.e, width of river = x + y = 4 [latex]\sqrt{3}[/latex]
15. The angle of elevation of top of a tower from a point on the groin 20 m away from the foot of the tower is 60°. The height of tower is

A. [latex]20 \frac{\sqrt{3}}{3}[/latex] m B. [latex]40 \frac{\sqrt{3}}{3}[/latex] m C. 20 m D. 20[latex]\sqrt{3}[/latex] m

Answer - Option D
Explanation -
tan 60º = [latex]\sqrt{3}[/latex] = [latex]\frac{n}{20}[/latex] =20 [latex]\sqrt{3}[/latex]

1. The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 7.5 m away from the wall. The length of the ladder is

A. 15 m B. 14.86 m C. 15. 64 m D. 15.8 m

Answer - Option A
Explanation -
According the question:
cos 60° = [latex]\frac{BC}{AC}[/latex]
[latex]\frac{1}{2}[/latex] = [latex]\frac{7.5}{AC}[/latex]
A = 15 m
2. If sinθ + cosθ =[latex]\sqrt{3}[/latex], then the value of [latex]\frac{3}{4}[/latex](tanθ + cotθ ) is

A. 1 B. [latex]\frac{3}{4}[/latex] C. [latex]\frac{3}{2}[/latex] D. 3

Answer - Option B
Explanation -
[latex]{sinθ + cosθ }^{2}[/latex] = [latex]{sin}^{2}θ + {cos}^{2}θ + 2cosθsinθ [/latex] = 3
cosθ. sinθ = 1, since [latex]{sin}^{2}θ + {cos}^{2}θ[/latex] = 1
[latex]\frac{3}{4}[/latex](tanθ + cotθ) = [latex]\frac{3}{4}[/latex][latex]\frac{{sin}^{2}θ + {cos}^{2}θ}{cosθsinθ}[/latex]
= [latex]\frac{3}{4}[/latex]
3. A ship is approaching a light home, 100 m high above the sea-level. The angle of depression of the ship as observed from the top of the light home, changes from 30° to 45°. The distance, in m, traveled by the ship during the period of observation , in m , is

A. 100([latex]\sqrt{3} + 1[/latex]) B. 100([latex]\sqrt{3} - 1[/latex]) C. ([latex]100 \sqrt{3} + 1[/latex]) D. ([latex]100 \sqrt{3} - 1[/latex])

Answer - Option B
Explanation -
X = [latex]100 \sqrt{3} - 100[/latex]
4. If [latex]\frac{secθ + tanθ }{secθ - tanθ}[/latex] = [latex]\frac{5}{3}[/latex] 0º < 0 < 90º then the value of cosecθ is

A. 2 B. [latex] \frac{\sqrt{15}}{4}[/latex] C. [latex] \frac {4}{\sqrt{15}}[/latex] D. 4

Answer - Option D
Explanation -
[latex]\frac{secθ + tanθ }{secθ - tanθ}[/latex] = [latex]\frac{5}{3}[/latex]
8 tanθ = 2 secθ
[latex]\frac{4 sinθ}{cosθ}[/latex] = [latex]\frac{1}{cosθ}[/latex] = sinθ = [latex]\frac{1}{4}[/latex]
cosecθ = 4
5. The length of the shadow of a pole is 90 m, when the sun's elevation is 30°.The length of the shadow of the pole is x meter when Sun's elevation is 60°. The value of x is

A. 15 [latex]\sqrt{3}[/latex] B. 30 C. 45 D. 20 [latex]\sqrt{3}[/latex]

Answer - Option B
Explanation -
tan 30º = [latex]\frac{h}{90}[/latex] = h = [latex] \frac {90} {\sqrt{3}}[/latex]
tan 60º = [latex]\frac{h}{x}[/latex] = x = [latex] \frac {90} {\sqrt{3} * \sqrt{3}}[/latex]
since tan 30º = [latex] \frac {1} {\sqrt{3}}[/latex]
and tan 60º = [latex]\sqrt{3}[/latex]
6. If 3 sinθ+ 4 cosθ = 5, then the value of 4 sinθ – 3cosθ is

A. -5 B. -2 C. 1 D. 0

Answer - Option D
Explanation -
3 sinθ+ 4 cosθ = 5
[latex]\frac{3}{5}[/latex] sin θ+ [latex]\frac{4}{5}[/latex] cos θ = 5,
comparing with sinθ . sin θ + cos θ – cos θ = 1
we get sin θ = [latex]\frac{3}{5}[/latex]
cos θ = [latex]\frac{4}{5}[/latex]
= 4 sinθ - 3 cosθ = 5
= 4 [latex]\frac{3}{5}[/latex] - 3 [latex]\frac{4}{5}[/latex] = 0
7. From the top of a tower 50 m high the angles of depression of the top and bottom of a pole are found to be 45° and 60° respectively. The height of the pole, in metre, is

A. 50 ([latex]\sqrt{3} - 1[/latex]) B. 50 ([latex]\sqrt{3} + 1[/latex]) C. 50([latex]3 - \sqrt{3} [/latex]) m D. 50([latex]6 \sqrt{3} - 1[/latex]) m

Answer - Option A
Explanation -
According the question:
tan 45º = 1 = [latex]\frac{x}{50 + h}[/latex]
x = 50 + h
tan 60º = [latex]\frac{x}{50}[/latex]
[latex]\sqrt{3}[/latex] * 50 = 50 + h
h = 50 ([latex]\sqrt{3} - 1[/latex])
since (tan 60º = [latex]\sqrt{3}[/latex])
8. If x = (1 + cotθ – cosecθ)(1 + tanθ + secθ),0º < 0 < 90º then the value of x is

A. -2 B. -1 C. 1 D. 2

Answer - Option D
Explanation -
(1 + cotθ – cosecθ)(1 + tanθ + secθ)
[latex](1 + \frac{cosθ}{sinθ} - \frac{1}{sinθ})(1 + \frac{sinθ}{cosθ} - \frac{1}{cosθ})[/latex]
= 2 + [latex]\frac{{sin}^{2}θ + {cos}^{2}θ - 1 }{sinθcosθ}[/latex]
= 2 + 0 = 2
9. The angle of elevation of the top of an unfinished tower at a point 20m away from its base is 45°. How much higher must the tower be raised so that its angle of elevation of the top at the same point be 60°

A. 20 [latex]\sqrt{3}[/latex] B. 20[latex]\sqrt{3}[/latex] + 1 m C. 20[latex](\sqrt{3} - 1)[/latex] m D. [latex]\frac{20 \sqrt{3} }{3} + 1[/latex] m

Answer - Option C
Explanation -
tan 45º = 1 = [latex]\frac{x}{20}[/latex]
x = 20
tan 60º = [latex]\sqrt{3}[/latex] = [latex]\frac{h + x}{20}[/latex]
h = 20 [latex]\sqrt{3}[/latex] - 20
= 20 [latex](\sqrt{3} - 1)[/latex]
10. The expression [latex]{tan}^{2}θ + {cot}^{2}θ - {sec}^{2}θ{cosec}^{2}θ [/latex] is equal to

A. 0 B. 1 C. -1 D. -2

Answer - Option D
Explanation -
[latex]{tan}^{2}θ + {cot}^{2}θ - {sec}^{2}θ{cosec}^{2}θ [/latex]
put θ= 45°, since all 'θ' values should give same value of expression
= 1 + 1 - [latex](\sqrt{2}^{2})(\sqrt{2}^{2})[/latex]
= 2 - 4 = -2
11. Two poles of equal height stand on either side of a roadway which is 100 m wide. At a point in the roadway between the poles, the elevation of the tops of the poles are 60°and 30°. the height of each pole, in meter is

A. 25 [latex]\sqrt{3}[/latex] B. 50 [latex]\sqrt {3}[/latex] C. 25 [latex](\sqrt{3} - 1)[/latex] D. 25 [latex](\sqrt{3} + 1)[/latex]

Answer - Option A
Explanation -
tan 60º = [latex]\sqrt{3}[/latex] = [latex]\frac{AB}{BC}[/latex]
BC = [latex]\frac{X}{\sqrt{3}}[/latex]
tan 30º = [latex]\frac {1} {\sqrt{3}}[/latex] = [latex]\frac{ED}{CD}[/latex]
CD = x [latex]\sqrt{3}[/latex]
[latex]\frac{A}{Q}[/latex] = [latex]\frac {x} {\sqrt{3}}[/latex] + x[latex]\sqrt{3}[/latex] = 100
x = 25 [latex]\sqrt{3}[/latex]
12. If [latex]{2sin}^{2}θ - 5sinθcosθ + {7cos}^{2}θ = 1 [/latex]then possible values of tanθ are

A. 2, 3 B. 1, 3 C. 2, [latex]\frac{5}{2}[/latex] D. 3, 4

Answer - Option A
Explanation -
[latex]{2sin}^{2}θ - 5sinθcosθ + {7cos}^{2}θ = 1 [/latex]
Divide by [latex]{cos}^{2}θ [/latex]
= [latex]{2 tan}^{2}θ - 5 tanθ + 4 [/latex]
= [latex]\frac{{sin}^{2}θ + {cos}^{2}θ}{{cos}^{2}θ}[/latex] = [latex]{tan}^{2}θ + 1[/latex]
= [latex]{tan}^{2}θ - tanθ + 6[/latex]
(tanθ - 3) (tanθ - 2) = 0
tanθ = 2 or 3
13. From the top of a tower 90 m high, the angles of depression of the top and bottom of a building are observed to be 30° and 60° respectively. The height of the building, in m, is

A. 40 B. 45 [latex]\sqrt{3}[/latex] C. 60 D. 40 [latex]\sqrt{3}[/latex]

Answer - Option C
Explanation -
tan 60º = [latex]\sqrt{3}[/latex] = [latex]\frac{90}{BC}[/latex]
= BC = [latex]\frac{90}{\sqrt{3}}[/latex]
tan 30º = [latex]\frac{1}{\sqrt{3}}[/latex] = [latex]\frac{90}{\frac {AE}{\sqrt{3}}}[/latex]
AE = 30
i.e, BE = 60 m
14. If cosθ – sinθ = l and tanθ = m [latex]{sec}^{2}θ[/latex], then the value of [latex]\frac{{I}^{2} + 2m}{2}[/latex] is

A. [latex]\frac{1}{2}[/latex] B. [latex]\frac{1}{4}[/latex] C. [latex]\frac{2}{4}[/latex] D. 1

Answer - Option A
Explanation -
cosθ– sinθ = l ; tanθ = m sec2θ
put θ = 45°
l = 0 ; m = [latex]\frac{1}{2}[/latex]
[latex]\frac{1 + 2m}{2}[/latex] = [latex]\frac{0 + 2 * \frac{1}{2}}{2}[/latex] = [latex]\frac{1}{2}[/latex]
15. A person is standing on the ground and flying a kite with a string of length 140 m at an angle of 30°. Another person is standing on the roof of a building 20 m high and is flying a kite at an angle of 45°. If both persons are on opposite sides of both the kites, the length (in m ) of the string that the second person must have so that the two kites meets,

A. 70 B. 60 [latex]\sqrt{2}[/latex] C. 50 [latex]\sqrt{2}[/latex] D. 50

Answer - Option C
Explanation -
sin 30° = [latex]\frac{1}{2}[/latex] = [latex]\frac{AB}{AC}[/latex]
AB = 70m
AF = (70 – 20) m = 50m
AE = 50 [latex]\sqrt{2}[/latex]