Tag: trigonometric identities

Questions Related to trigonometric identities

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

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