Tag: trigonometry

Questions Related to trigonometry

Find sec $\theta$, if $\theta$ be an angle in standard position with (x, y) a point on the terminal side of $\theta$ and r = $\sqrt{x^{2}+y^{2}}\neq 0$.

maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

  1. $\dfrac{y}{r}$

  2. $\dfrac{r}{y}$

  3. $\dfrac{x}{r}$

  4. $\dfrac{r}{x}$

Show answer
Correct Option: D
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$x^{2}+y^{2}=r^{2}$

So, sec $\theta$ = $\frac{hypotenuse}{adjacent\space\ side }$= $\dfrac{r}{x}$

sec $\theta$ = $\dfrac{r}{x}$

So, option D is correct.

Find the exact value of sin $\theta$ trigonometric functions for the angle formed when the terminal side passes through (6, 8).

maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

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

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

  3. $\dfrac{10}{5}$

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

Show answer
Correct Option: A
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, $\sin \theta$ = $\dfrac{opposite \space\ side }{hypotenuse}$

= $\dfrac{8}{10}$ = $\dfrac{4}{5}$

So, option A is correct.

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

maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

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

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

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

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

Show answer
Correct Option: C
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, $\csc \theta$ = $\dfrac{hypotenuse}{opposite \space\ side}$

$\csc \theta$ = $\dfrac{5}{3}$

So, option C is correct.

Find cot $\theta$, if $\theta$ be an angle in standard position with (x, y) a point on the terminal side of $\theta$ and r = $\sqrt{x^{2}+y^{2}}\neq 0$.

maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

  1. $\dfrac{y}{r}$

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

  3. $\dfrac{x}{r}$

  4. $\dfrac{r}{x}$

Show answer
Correct Option: B
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$x^{2}+y^{2}=r^{2}$
So, cot $\theta$ = $\dfrac{adjacent\space\ side}{opposite\space\ side}$= $\dfrac{x}{y}$


cot $\theta$ = $\dfrac{x}{y}$

So, option B is correct.

Find cosec $\theta$, if $\theta$ be an angle in standard position with (x, y) a point on the terminal side of $\theta$ and r = $\sqrt{x^{2}+y^{2}}\neq 0$.

maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

  1. $\dfrac{y}{r}$

  2. $\dfrac{r}{y}$

  3. $\dfrac{x}{r}$

  4. $\dfrac{r}{x}$

Show answer
Correct Option: B
Explanation:

By using Pythagoras theorem, we will find the value of hypotenuse.
$x^{2}+y^{2}=r^{2}$
So, cosec $\theta$ = $\dfrac{hypotenuse}{opposite\space\ side }$= $\dfrac{r}{y}$

cosec $\theta$ = $\dfrac{r}{y}$

So, option B is correct.


lf the distance between the points $(a \cos\theta, a \sin\theta), (a \cos\phi, a \mathrm{sin} \phi)$ is $2\mathrm{a}$ then $\theta=$.

maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

  1. $2n\pi\pm\pi+\phi, n\in z$

  2. $n\displaystyle \pi\pm\frac{\pi}{2}+\phi, n\in z$

  3. $n\pi-\phi, n\in z$

  4. $2n\pi+\phi, n\in z$

Show answer
Correct Option: A
Explanation:
Given points $(a=cos\theta, a\sin\theta), (a\cos\phi, a\sin \phi)$

distance $2a$

distance between $(x _{1}, y _{1}), (x _{2}, y _{2})$

$\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{2})^{2}}$

So, 
$\sqrt{(a\cos\theta-a\cos\theta)^{2}+(a\sin\phi-a\sin\theta)^{2}}=2a$

Squaring on both sides

$\Rightarrow a^{2}(\cos^{2}\theta+\cos^{2}\phi-2\cos\theta\cos\phi)+a(\sin^{2}\phi+\sin^{2}\theta-2\sin\theta-\sin\phi)=4a^{2}$

$\Rightarrow a^{2}(\sin^{2}\theta+\cos^{2}\theta+\sin^{2}\theta+\cos^{2}\theta-2(\sin\theta-\sin\phi+\cos\theta\cos\phi))=4a^{2}$

$\Rightarrow a^{2}(1+1-2(\cos(\theta-\phi))=4a^{2}$

$\Rightarrow 2-2\cos(\theta-\phi)=4$

$\Rightarrow 2\cos(\theta-\phi)=-2$

$\Rightarrow \cos(\theta-\phi)=-1=\cos(\pi)$

$\Rightarrow \theta-\phi=2n\pi\pm \pi, n\in z$

$\Rightarrow \boxed{\theta=2n\pi\pm\pi+\phi, n\in z}$

Find the exact value of $\dfrac{\sin \theta}{\sec \theta}$ trigonometric functions for the angle formed when the terminal side passes through $(3, 4)$.

maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

  1. $\dfrac{25}{12}$

  2. $\dfrac{12}{5}$

  3. $\dfrac{12}{25}$

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

Show answer
Correct Option: C
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, $\dfrac{\sin\theta}{\sec\theta}$ = $\dfrac{opposite \space\ side \times adjacent \space side}{hypotenuse \times hypotenuse}$
= $\dfrac{3\times 4}{5\times 5}$
=$\dfrac{12}{25}$

So, option C is correct.

If $\tan 4x+\tan 5x-\tan 9x=k\tan 4x\tan 5x\tan 9x$ then $k=$

trigonometric equations trigonometric functions trigonometry maths

  1. $1$

  2. $-1$

  3. $ \pm 1$

  4. $2$

Show answer
Correct Option: B
Explanation:

$\begin{array}{l}\tan 9x = \tan (4x + 5x) = \dfrac{{\tan 4x + \tan 5x}}{{1 - \tan 4x\tan 5x}}\ \Rightarrow \tan 9x - \tan 4x\tan 5x\tan 9x = \tan 4x + \tan 5x\ \Rightarrow \tan 4x + \tan 5x - \tan 9x =  - \tan 4x\tan 5x\tan 9x\\therefore k =  - 1\end{array}$

State true or false $\tan(\dfrac{\pi}{4} + \theta) - \tan(\dfrac{\pi}{4} -\theta) = 2\tan\theta$

trigonometric equations trigonometric functions trigonometry maths

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Using $\tan (A+B)=\cfrac {\tan A+\tan B}{1-\tan A \tan B}$

$\Rightarrow \tan (A-B)=\cfrac {\tan A-\tan B}{1+\tan A+\tan B}$
$\tan \left( \cfrac {\pi}{4}+\theta\right)-\tan \left(\cfrac {\pi}{4}-\theta\right)$
$\Rightarrow A=\cfrac {\pi}{4}, B=\theta$
$\Rightarrow \cfrac {\tan \cfrac {\pi}{4}+\tan\theta}{1-\tan \cfrac {\pi}{4}\tan \theta}-\cfrac {\tan \cfrac {\pi}{4}-\tan \theta}{1+\tan \cfrac {\pi}{4}\tan \theta}$
$\because \tan \cfrac {\pi}{4}=1$
$\Rightarrow \cfrac {1+\tan\theta}{1-\tan \theta}-\cfrac {1-\tan \theta}{1+\tan \theta}$
$\Rightarrow \cfrac {(1+\tan\theta)^2-(1-\tan\theta)^2}{(1-\tan\theta)(1+\tan\theta)}$
$\therefore a^2-b^2=(a+b)(a-b)$
$\Rightarrow \cfrac {1+\tan^2\theta+2\tan\theta-1-\tan^2\theta+2\tan\theta}{1-\tan^2\theta}$
$\Rightarrow \cfrac {4\tan\theta}{1-\tan^2\theta}$
$\because \tan2\theta=\cfrac {2\tan\theta}{1-\tan^2\theta}$
$=2\tan2\theta$.

State true or false

$\tan{ 18 }^{ 0 }+\tan27^{ 0 }+\tan{ 18 }^{ 0 }\cdot \tan27^{ 0 }=1$

trigonometric equations trigonometric functions trigonometry maths

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

$tan180+tan270+tan180+tan270=1$

$0.32491969 + 0.50952544+0.32491969 + 0.50952544 = 1$

$1.66889026 = 1$

The left side $1.66889026 $ not equal to rightside$ 1$.

Which means the given statement is false.