Tag: trigonometric ratios of acute angles

Questions Related to trigonometric ratios of acute angles

$\sin ^{ 8 }{ \theta  } -\cos ^{ 8 }{ \theta  } -\left( \sin ^{ 2 }{ \theta  } -\cos ^{ 2 }{ \theta  }  \right) \left( 1-\sin ^{ 2 }{ \theta  }  \right) $=0

  1. True

  2. False


Correct Option: B
Explanation:

$\sin^4 \theta+\cos^4 \theta=(\sin^2 \theta+\cos^2 \theta)^2-2\sin^2 \theta\cos^2 \theta=1-2\sin^2 \theta\cos^2 \theta$

$\sin^4 \theta-\cos^4 \theta=(\sin^2 \theta-\cos^2 \theta)(\sin^2 \theta+\cos^2 \theta)=\sin^2 \theta-\cos^2 \theta$
$\sin^8\theta-\cos^8\theta-(\sin^2 \theta-\cos^2 \theta)(1-\sin^2 \theta)=(\sin^4 \theta+\cos^4 \theta)(\sin^4 \theta-\cos^4 \theta)-(\sin^2 \theta-\cos^2 \theta)(1-\sin^2 \theta)$
                                                                                 $=(1-2\sin^2 \theta\cos^2 \theta)(\sin^2 \theta-\cos^2 \theta)-(\sin ^2 \theta-\cos^2\theta)(1-\sin^2 \theta)$
                                                                                 $=(\sin^2 \theta-\cos^2 \theta)(\sin^2 \theta)(1-2\cos^2 \theta)$
So the given relation is $\text{False}$

If $tan x + cot x = 2$, then $sin^{2n}x+cos^{2n}x=$

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

  2. $2^n$

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

  4. $\dfrac{1}{2^{n-1}}$


Correct Option: D
Explanation:

Given $\tan x+\cot x=2$

$\implies \tan x+\dfrac{1}{\tan x}=2$
$\implies \tan^2 x-2\tan x+1=0$
$\implies (\tan x-1)^2=0$
$\implies \tan x=1\implies x=\dfrac{\pi}{4}$
$\sin^{2 n} x+\cos^{2 n} x=\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2 n}+\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2 n}=\dfrac{1}{2^n}+\dfrac{1}{2^n}=\dfrac{2}{2^n}=\dfrac{1}{2^{n-1}}$

If $\tan x =\dfrac{3}{4} , \pi < x < \dfrac{3\pi}{2} $ find value of $\sin\dfrac{x}{2} , \cos\dfrac{x}{2} , \tan \dfrac{x}{2}$

  1. cos x/2= 3/2, sin x/2 =-3/2, tan x/2= -1/√10

  2. cos x/2= 3/√10, sin x/2 =-3/2, tan x/2= -1/√10

  3. cos x/2= -1/√10, sin x/2 = 3/√10, tan x/2= -3

  4. cos x/2= 3/√10, sin x/2 =-2/3, tan x/2= -1/√10


Correct Option: C
Explanation:
Given 

$ \pi < x < \dfrac {3\pi} 2$ 

$ \implies \dfrac \pi 2 < \dfrac x2 < \dfrac {3\pi }4 $ 

$ \implies \cos \dfrac x2 <0 , \sin \dfrac x2 >0 $ 

$ \tan x =\dfrac 34 $ 

$ \dfrac {2 \tan \dfrac x2 }{1-\tan ^2 \dfrac x2 }=\dfrac 34$

$ 8 \tan \dfrac x2 =3-3\tan ^2 \dfrac x2 $

$ 3\tan ^2 \dfrac x2 +8\tan \dfrac x2 -3=0 $ 

$ 3\tan ^2 \dfrac x2 +9\tan \dfrac x2 - \left(\tan \dfrac x2 +3 \right)=0 $ 

$ \tan \dfrac x2 =-3,\dfrac 13 $ 

As $ tan \dfrac x2 <0 \implies \tan \dfrac x2 =-3$ 

$ \cos \dfrac x2 =\dfrac {-1}{\sqrt {(-1)^2+3^2}}=\dfrac {-1}{\sqrt {10}}$ 

$ \sin \dfrac x2 =\dfrac {3}{\sqrt {(-1)^2+3^2}}=\dfrac {3}{\sqrt {10}}$ 

A boat takes $19$ hours for travelling downstream from point $A$ to point $B$ and coming back to a point $C$ midway between $A$ and $B$. If the velocity of the stream is $4$ $kmph$ and the speed of the boat in still water is $14 \mathrm { kmph}, $ what is the distance between $A$ and $B$?

  1. $160km$

  2. $180km$

  3. $200km$

  4. $220km$


Correct Option: B
Explanation:


Speed in downstream =  $(14 + 4) km/hr = 18 km/hr ;$

Speed in upstream =  $(14 – 4) km/hr = 10 km/hr. $

Let the distance between A and B be $x$ km. Then,

$\frac{x}{18} + \frac{x}{2}\times\frac{1}{10}$ = 19 

$\Rightarrow \frac{x}{18} + \frac{x}{20}$ = 19 

$\Rightarrow$ $x = 180 km.$

If $\tan \theta+\left(\dfrac {\pi}{2}+\theta\right)=0$ then the most general value of $\theta$ is (where $n\ \in\ Z$)

  1. $n\ \pi \pm \dfrac {\pi}{4}$

  2. $2n\ \pi \pm \dfrac {\pi}{4}$

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

  4. $\dfrac {n\pi}{2}+(-1)^{n} ,\dfrac {\pi}{4}$


Correct Option: A

$\sin\ (45^{o}+\theta)-\cos\ (45^{o}-\theta)$ is equal to

  1. $2\cos \theta$

  2. $0$

  3. $2\sin \theta$

  4. $1$


Correct Option: A

One angle of a triangle is $\displaystyle \frac{2x}{3}$ grades another is $\displaystyle \frac{3x}{2}$ degrees, whilst the third is $\displaystyle \frac{2\pi x}{75}$ radians ; express them all in degrees.

  1. ${ 55 }^{ o },\quad { 28 }^{ o }\quad &amp; \quad { 97 }^{ o }\$

  2. ${ 65 }^{ o },\quad { 22 }^{ o }\quad &amp; \quad { 93 }^{ o }\$

  3. $\{ 60 }^{ o },\quad { 24 }^{ o }\quad &amp; \quad { 96 }^{ o }\$

  4. ${ 70 }^{ o },\quad { 15 }^{ o }\quad &amp; \quad { 95 }^{ o }\$


Correct Option: C

The angles of elevation of the top of the tower from two points at distances '$a$' and '$b$' from the base and in the same straight line with it complementary The height of the tower is

  1. $a + b$

  2. $\displaystyle \sqrt{ab}$

  3. $\displaystyle a\times b$

  4. $\displaystyle a\sqrt{b}$


Correct Option: B
Explanation:

AP=a,AQ=b
$\displaystyle \tan \theta =\frac{h}{a}$....(i)
$\displaystyle \tan \left ( 90-\theta  \right )=\frac{h}{b}$....(ii)
$\displaystyle \Rightarrow \cot \theta =\frac{h}{b}$
$\displaystyle \Rightarrow \tan \theta \times \cot \theta = \frac{h}{a}\times \frac{h}{b}=1$
$\displaystyle \Rightarrow h^{2}=ab$
or $\displaystyle h=\sqrt{ab}$

What is the exact value of $cos \theta$ trigonometric functions for the angle formed when the terminal side passes through $(6, 8)$?

  1. $\dfrac{3}{5}$

  2. $\dfrac{6}{5}$

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

  4. $\dfrac{8}{5}$


Correct Option: C
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$6^{2}+8^{2}=c^{2}$
$36 + 64 = c^{2}$
$c = 10$
So, $\cos \theta$ = $\dfrac{opposite \space\ side }{hypotenuse}$ = $\dfrac{8}{10}$

$\cos \theta$ = $\dfrac{4}{5}$

So, option C is correct.

Determine the exact value of $\cos \theta$ trigonometric functions for the angle formed when the terminal side passes through $(3, 4).$

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

  2. $\dfrac{3}{5}$

  3. $\dfrac{2}{5}$

  4. $\dfrac{1}{5}$


Correct Option: A
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$3^{2}+4^{2}=c^{2}$
$9 + 16 = c^{2}$
$c = 5$
So, $\cos\theta$ = $\dfrac{adjacent \space\ side}{hypotenuse}$

$\cos \theta$ = $\dfrac{4}{5}$

So, option A is correct.