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Using trigonometric tables - class-XI

Description: using trigonometric tables
Number of Questions: 49
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Tags: trigonometric identities trigonometry maths
Attempted 0/48 Correct 0 Score 0

The value of ${ cosec }^{ 2 }{ 51 }^{ 0 }-{ cot }^{ 2 }{ 51 }^{ 0 }$ is 

  1. 1

  2. -1

  3. 0

  4. $\frac { 1 }{ 2 } $


Correct Option: A
Explanation:

We know,

$1 + \cot^2\theta = cosec^2 \theta$

$\therefore cosec^2\theta - \cot^2 \theta = 1$

Thus, $cosec^2  51^o - \cot^251^o = 1$

Hence, option a is correct.

Choose the correct option for the following statement.

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

  1. The given statement is true 

  2. The given statement is false

  3. Incomplete information

  4. None of the above.


Correct Option: A
Explanation:

By definition,

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
Therefore, the given statement is true.

$\sin { { 30 }^{ o }= } $

  1. $\dfrac {1}{2}$

  2. $-\dfrac {1}{2}$

  3. $\dfrac {\sqrt {3}}{2}$

  4. $-\dfrac {\sqrt {3}}{2}$


Correct Option: A
Explanation:

As we know that 

$\sin 30^{\circ}=\dfrac{1}{2}$

The value of $\tan 7\dfrac{1}{2}^{o}$ is equal to

  1. $\sqrt{6}+\sqrt{3}+\sqrt{2}-2$

  2. $\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

  3. $\sqrt{6}-\sqrt{3}+\sqrt{2}+2$

  4. $\sqrt{6}-\sqrt{3}-\sqrt{2}-2$


Correct Option: A

General solution of $cotx+tanx=2cosecx$ is

  1. $2n\pi\pm\dfrac{2\pi}{3},n\inZ$

  2. $2n\pi\pm\dfrac{4\pi}{3},n\in Z$

  3. $2n\pi\pm\dfrac{5\pi}{3},n\in Z$

  4. $2n\pi\pm\dfrac{\pi}{3},n\in Z$


Correct Option: A

If $\cos A=\cos $ and $\sin A=\sin B$ then 

  1. $A+B=0$

  2. $A=B$

  3. $A+B=2n \pi$

  4. $A=B+2n \pi$


Correct Option: A

$\left( 1-\dfrac { \cos { 61^{o} }  }{ \cos { 1^{o} }  }  \right) \left( 1-\dfrac { \cos { 62^{o} }  }{ \cos { 2^{o} }  }  \right) \left( 1-\dfrac { \cos { 63^{o} }  }{ \cos { 3^{o} }  }  \right) .......\left( 1-\dfrac { \cos { 119^{o} }  }{ \cos { 59^{o} }  }  \right) $

  1. $-1$

  2. $1$

  3. $2$

  4. $0$


Correct Option: A

If $\tan \theta =\dfrac {\cos 9^{o}+\sin 9^{o}}{\cos 9^{o}-\sin 9^{o}}$, then the value of $\theta$ is

  1. $9^{o}$

  2. $54^{o}$

  3. $18^{o}$

  4. $None\ of\ these$


Correct Option: A

The value of $\tan \theta .\tan (\theta +60^{o})+\tan \theta \tan (\theta -60^{o})+\tan (\theta +60^{o}).\tan (\theta -60^{o})+3$ is  

  1. $0$

  2. $1$

  3. $-1$

  4. $None\ of\ these$


Correct Option: A

The value of $\sin 75^{o}$ is

  1. $\dfrac {2-\sqrt {3}}{\sqrt {2}}$

  2. $\dfrac {\sqrt {3}+1}{2\sqrt {2}}$

  3. $\dfrac {\sqrt {3}-1}{2\sqrt {2}}$

  4. $\dfrac {\sqrt {3}+1}{\sqrt {2}}$


Correct Option: A

$\dfrac{\cos{20}^{o}+8\sin{70}^{o}\sin{50}^{o}\sin{10}^{o}}{{\sin}^{2}{80}^{0}}$ is equal to:

  1. $1$

  2. $2$

  3. $\dfrac{3}{4}$

  4. $\none$


Correct Option: A

The value of expression $\dfrac { 2\left( \sin{ 1 }^{ o }+\sin{ 2 }^{ o }+\sin{ 3 }^{ o }+.....+\sin{ 89 }^{ o } \right)  }{ 2\left( \cos{ 1 }^{ o }+\cos{ 2 }^{ o}+......+\cos{ 44 }^{ o } \right) +1 }$ equals

  1. $\sqrt{2}$

  2. $1/\sqrt{2}$

  3. $1/2$

  4. $0$


Correct Option: A

${\cos}^{2}{73}^{o}+{\cos}^{2}{47}^{o}+\cos{73}^{o}\cos{47}^{o}=.$

  1. $\dfrac{3}{4}$

  2. $-\dfrac{3}{4}$

  3. $\dfrac{4}{3}$

  4. $-\dfrac{4}{3}$


Correct Option: A

$\dfrac { \cos{ 13 }^{ o }-\sin{ 13 }^{ o } }{ \cos{ 13 }^{ o }+\sin{ 13 }^{ o } } +\dfrac { 1 }{ \cot{ 148 }^{ o } }$ is equal to

  1. $1$

  2. $-1$

  3. $0$

  4. $\dfrac { 1 }{ 2 } $


Correct Option: C
Explanation:

$\dfrac{\cos 13 - \sin 13}{\cos 13 + \sin 13} + \dfrac{1}{\cot 148}$


$=\dfrac{\cos 13 (1 - \tan 13)}{\cos 13 (1 + \tan 13)} + \dfrac{1}{\cot (180 - 32)}$


$=\dfrac{\tan 45 - \tan 13}{1 + \tan 45 \tan 13} + \dfrac{1}{(-\cot 32)}$

$=\tan (45 - 13) - \tan 32$

$=\tan (32) - \tan 32$

$=0$

The value of $\sqrt { 3 } tan{ 10 }^{ 0 }+\sqrt { 3 } tan{ 20 }^{ 0 }+tan{ 10 }^{ 0 }tan{ 20 }^{ 0 }$ is ___________.

  1. $-1$

  2. $0$

  3. $1$

  4. $2$


Correct Option: C
Explanation:

$\tan (30) = \tan (20 + 10)$


$\dfrac{1}{\sqrt{3}} = \tan 30 = \dfrac{\tan 20 + \tan 10}{1 - \tan 20 \tan 10}$


$1 - \tan 20 \tan 10 = \sqrt{3} (\tan 20 + \tan 10)$

$\sqrt{3} \tan 20 + \sqrt{3} \tan 10 + \tan 10 \tan 20 = 1$

If $sin(A-B)=\frac { 1 }{ 2 } ,cos(A+B)=\frac { 1 }{ 2 } ,{ 0 }^{ 0 }<A+B\le { 90 }^{ 0 }$ then A =

  1. $15^{ 0 }$

  2. $45^{ 0 }$

  3. $90^{ 0 }$

  4. $30^{ 0 }$


Correct Option: A

The value of $cos^2 10^o 15^o + cos^2 20^o +...... + cos^2 365^O$

  1. 34

  2. 36

  3. 35

  4. 37/2


Correct Option: A

$16 \cos^6 10^o - 24 \cos^4 10^o + 9 \cos^2 10^o$ is equal to

  1. $\dfrac{1}{4}$

  2. $\dfrac{3}{4}$

  3. $\dfrac{1}{2}$

  4. $1$


Correct Option: A

If $(1+\tan 1^{o})(1+\tan 2^{o})(1+\tan 3^{o})....(1+\tan 45^{o})=2^{n}$, then $n$ is equal to 

  1. $21$

  2. $24$

  3. $23$

  4. $22$


Correct Option: A

Values of : $sin{ 10 }^{ 0 }sin{ 50 }^{ 0 }sin{ 60 }^{ 0 }sin{ 70 }^{ 0 }$ is

  1. $\cfrac { 3 }{ 16 } $

  2. $\cfrac { 5 }{ 16 } $

  3. $\cfrac { \sqrt { 3 } }{ 16 } $

  4. $\cfrac { \sqrt { 5 } }{ 16 } $


Correct Option: C
Explanation:

As we know that

$\sin A\sin(60^{\circ}-A)\sin (60^{\circ}+A)=\dfrac{1}{4}\sin 3 A$
Put $A=10^{\circ}$
So $\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\dfrac{1}{4}\sin 30^{\circ}=\dfrac{1}{8}$
So $\sin 10^{\circ}\sin 60^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\sin 60^{\circ}(\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ})=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{8}=\dfrac{\sqrt{3}}{16}$

$sin^21^0+sin^22^0+sin^23^0+....+sin^290^0$

  1. 0

  2. 1

  3. 89/2

  4. 91/2


Correct Option: A

$\frac { cos{ 70 }^{ \circ  } }{ sin{ 20 }^{ \circ  } } +\frac { cos{ 59 }^{ \circ  } }{ sin{ 31 }^{ \circ  } } -8{ sin }^{ 2 }{ 30 }^{ \circ  }$

  1. $1$

  2. $-1$

  3. $0$

  4. $2$


Correct Option: A

The value of $(4 \, cos^2 9^o - 1) (4 cos^2 27^o - 1) (4 \, cos^2 81^o - 1) (4 \, cos^2 243^o - 1) $ is 

  1. 1

  2. -1

  3. 2

  4. none of these


Correct Option: A

$sin{ 40 }^{ \circ  }{ 35 }^{ | }cos{ 19 }^{ \circ  }{ 25 }^{ | }+cos{ 40 }^{ \circ  }{ 35 }^{ | }sin{ 19 }^{ \circ  }{ 25 }^{ |= }$

  1. 1

  2. $\frac { \sqrt { 3 } }{ 2 } $

  3. $0$

  4. $-1$


Correct Option: A

$\sqrt {3}\csc 20^{o}-\sec 20^{o}$ is equal to

  1. $2$

  2. $2\sin 20^{o}\csc 40^{o}$

  3. $4$

  4. $4\sin 20^{o}\csc 40^{o}$.


Correct Option: B

$tan{ 40 }^{ \circ  }+tan{ 80 }^{ \circ  }-\sqrt { 3 } tan{ 40 }^{ \circ  }tan{ 80 }^{ \circ  }=$

  1. $\sqrt { 3 } $

  2. $\sqrt { -3 } $

  3. $\frac { 1 }{ \sqrt { 3 } }$

  4. $\frac { -1 }{ \sqrt { 3 } }$


Correct Option: A

$\sin^{-1}\left(\sin 100\right)+\cos^{-1}\left(\cos 100\right)+\tan^{-1}\left(\tan 100\right)+\cot^{-1}\left(\cot 100\right)$ equals to 

  1. $100-31\pi$

  2. $100-32\pi$

  3. $200-63\pi$

  4. $200+63\pi$


Correct Option: A

$\dfrac{tan225 cot81cot69}{cot261 + tan21} $ = 

  1. 1

  2. $\dfrac{1}{\sqrt{2}}$

  3. $\sqrt{3}$

  4. $\dfrac{1}{\sqrt{3}}$


Correct Option: A

Find the value of, $\dfrac {4}{3}\cot^{2}30^{o}+\cot^{2}60^{o}-2\csc ^{2}60^{o}-\dfrac {3}{4}\tan^{2}30^{o}$

  1. $10/3$

  2. $11/3$

  3. $4$

  4. $none\ of\ these$


Correct Option: D
Explanation:

$\dfrac { 4 }{ 3 } { \cot }^{ 2 }30+{ \cot }^{ 2 }60-2{ csc }^{ 2 }60-\dfrac { 3 }{ 4 } { \tan }^{ 2 }30$

$\Rightarrow \dfrac { 4 }{ 3 } { \left( \sqrt { 3 }  \right)  }^{ 2 }+{ \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  }^{ 2 }-2\times { \left( \dfrac { 2 }{ \sqrt { 3 }  }  \right)  }^{ 2 }-\dfrac { 3 }{ 4 } \times { \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  }^{ 2 }$
$\Rightarrow 4+\dfrac { 1 }{ 3 } -\dfrac { 8 }{ 3 } -\dfrac { 1 }{ 4 } $
$\Rightarrow \dfrac { 15 }{ 4 } -\dfrac { 7 }{ 3 } $
$=\dfrac { 17 }{ 12 } $
None of these.

$ \sqrt { 3 } \ cosec 20 ^ { \circ } - \sec 20 ^ { \circ }$  is equal to :

  1. $2$

  2. $2 \sin 20 ^ { \circ } / \sin 40 ^ { \circ }$

  3. $4$

  4. $4 \sin 20 ^ { \circ } / \sin 40 ^ { \circ }$


Correct Option: A

$Tan \,25^o.Tan \,31^o + Tan \,31^o.Tan \,34^o  + Tan \,34^o .Tan \,25^o =$

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: A

Let $x=\sin 1^{o}$, then the value of expression
$\dfrac {1}{\cos 0^{o}\cos 1^{o}}+\dfrac {1}{\cos 1^{o}\cos 2^{o}}+\dfrac {1}{\cos 2^{o}\cos 3^{o}}+....+\dfrac {1}{\cos 44^{o}\cos 45^{o}}=$

  1. $x$

  2. $\dfrac {1}{x}$

  3. $\dfrac {\sqrt {2}}{x}$

  4. $\dfrac {x}{\sqrt {2}}$


Correct Option: A

$\dfrac{\cos (45^0+A)-\cos(45^0-A)}{\sin(120^0+A)-\sin(120^0-A)}=?$

  1. $2$

  2. $\sqrt{2}$

  3. $2\sqrt{2}$

  4. $2/\sqrt{2}$


Correct Option: A

if $A=30$ then the value of $\cos 2A$ is

  1. $1$

  2. $0$

  3. $1/2$

  4. $\surd {3}/2$


Correct Option: A

$(1+\tan 5^{o})(1+\tan 10^{o})(1+\tan 15^{o}).....(1+\tan 45^{o})$ is equal to

  1. $!$

  2. $32$

  3. $2$

  4. $0$


Correct Option: A

$sin 12^o\sin\ 24^o\sin\ 48^o\sin\ 84^o$=

  1. $\cos 20^o\cos 40^o\cos 60^o\cos 80^o$

  2. $\sin 20^o\sin 40^o\sin 60^o\sin 80^o$

  3. $\dfrac{3}{15}$

  4. $\dfrac{5}{16}$


Correct Option: A

$\tan 5\tan 25\tan 30\tan 65 \tan 85$ is equal to

  1. $3$

  2. $\surd {3}$

  3. $1$

  4. $\dfrac {1}{\surd {3}}$


Correct Option: A

$\tan 20 ^ { \circ } + \tan 40 ^ { \circ } + \sqrt { 3 } \tan 20 ^ { \circ } \tan 40 ^ { \circ }$  is equal to

  1. $\dfrac { \sqrt { 3 } } { 2 }$

  2. $\dfrac { \sqrt { 3 } } { 4 }$

  3. $\sqrt { 3 }$

  4. $1$


Correct Option: A

If $\alpha =685^o$, then $(\cos\alpha -\sin\alpha)$ is equivalent to?

  1. $-\cos 35^o-\sin 35^o$

  2. $-\cos 35^o +\sin 35^o$

  3. $\cos 35^o-\sin 35^o$

  4. $\cos 35^o+\sin 35^o$


Correct Option: A

The distance between $A ( \cos \theta , \sin \theta )$ and $B ( - \sin \theta , \cos \theta )$ is

  1. 1

  2. $2 + 2 \sin \theta$

  3. $1 + \sin \theta$

  4. $\sqrt { 2 }$


Correct Option: D
Explanation:

Given $A(\cos \theta,\sin \theta)$ and $B(-\sin \theta,\cos \theta)$

Distance between $AB=\sqrt{(\cos \theta+\sin \theta)^2+(\sin \theta-\cos \theta)^2}=\sqrt{\cos^2 \theta+\sin ^2 \theta+2\sin \theta\cos \theta+\cos^2\theta+\sin ^2 \theta-2\sin \theta\cos \theta}=\sqrt{1+1}=\sqrt{2}$

$\dfrac  { 1 tan^{ 1 } 45^{ 2 }}{ 1 tan^{ 1 } 45^{ 0 } }$

  1. $tan 90^{ 0 }$

  2. $sin 45^{ 0 }$

  3. $sin 90^{ 0 }$

  4. $cos 90^{ 0 }$


Correct Option: A

$\displaystyle3\tan^2{30^\circ}+\frac{4}{3}\cos^2{30^\circ}-2\sin^2{45^\circ}-\frac{1}{3}\sin^2{60^\circ}$ is equal to__________________.

  1. $\displaystyle\frac{1}{4}$

  2. $\displaystyle\frac{3}{4}$

  3. $1$

  4. $0$


Correct Option: B
Explanation:

Given 


$3\tan ^230+\dfrac 43\cos ^230-2\sin ^245-\dfrac 13\sin ^260$

$=3\left(\dfrac 1{\sqrt 3}\right)^2+\dfrac 43\left(\dfrac {\sqrt 3}2 \right)^2-2\left(\dfrac 1{\sqrt 2}\right)^2-\dfrac 13\left(\dfrac {\sqrt 3}2\right)^2$

$=1+1-1-\dfrac 14$

$= \dfrac 34$

The angle measuring $\displaystyle \frac{\pi ^{c}}{4}$ when expressed in centesimal system is ___ 

  1. $\displaystyle 50^{g}$

  2. $\displaystyle 60^{g}$

  3. $\displaystyle 75^{g}$

  4. $\displaystyle 100^{g}$


Correct Option: A
Explanation:

In $\text{Centesimal System}$, an angle is measured in grades, minutes and seconds.
Given angle $ = \dfrac {{\pi}^c}{4} = \dfrac {{180}^{0}}{4} = {45}^{0} $

We know that $ {1}^{0} = {(\dfrac {10}{9})}^{g} $


$ \Rightarrow {45}^{0} = {\dfrac {10}{9}} \times 45^{g}  ={50}^{g} $

$\displaystyle 30^{\circ}$ in centesimal measure is _____

  1. $\displaystyle \frac{50^{g}}{3}$

  2. $\displaystyle \frac{100^{g}}{3}$

  3. $\displaystyle \frac{160^{g}}{3}$

  4. $\displaystyle \frac{200^{g}}{3}$


Correct Option: B
Explanation:

In $\text{Centesimal System}$, an angle is measured in grades, minutes and seconds.
In centesimal system, we know that $ {1}^{0} = {(\dfrac {10}{9})}^{g} $
$ \Rightarrow  {30}^{0} =  {(\dfrac {10}{9})}^{g} \times 30 = {(\dfrac {100}{3})}^{g} $

When the sun is $30^o$ above the horizon, what is the length of the shadow cast by a building $40$ m high?

  1. $50.23$ m

  2. $70.24$ m

  3. $68.25$ m

  4. $69.28$ m


Correct Option: D
Explanation:

$\tan 30^o = \dfrac{40m}{shadow}$

Shadow $= \dfrac{40}{\tan 30^o}$

$= \dfrac{40}{\frac{1}{\sqrt{3}}}$= $\dfrac{40}{0.57735}$

The shadow of the building is $69.28$ m.

So, option D is correct.

From the tower $30$ m above the sea, the angle of depression of a boat is $68^o$. How far is the boat from the tower?

  1. $12.12$ m

  2. $11.11$ m

  3. $10.10$ m

  4. $9.99$ m


Correct Option: A
Explanation:
$\tan 68^o = \dfrac{30}{distance}$

Distance $=\dfrac{30}{\tan 68^o}$

Distance = $\dfrac{30}{2.475087}$

Distance $= 12.12$ m

So, option A is correct.

When the sun is $50^o$ above the horizon, how long is the shadow cast by a building $16$ m high?

  1. $23$ m

  2. $13.42$ m

  3. $43.42$ m

  4. $23.42$ m


Correct Option: B
Explanation:

$\tan 50^o = \dfrac{16m}{shadow}$

Shadow $= \dfrac{16}{\tan 50^o}$= $\dfrac{16}{1.191754}$= $13.42$

The shadow of the building is $13.42$ m

So, option B is correct.

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