Question

3.
If tan 0 = 2, find the values of other
trigonometric ratios.​

Answer 1

Note:

let theta=a

Given

Tana=2

To find

Values of other trigonometric functions

Explanation

Consider

$\bigtriangleup ABC,$

we know that

$\boxed{\bf tan=\dfrac{opposite}{adjacent}}$

tana=2/1

opposite=2

Adjacent=1

Hypotenuse ²=opposite ²+adjacent²

H²=2²+1²

H=√5 units

$\boxed{\bf sin=\dfrac{opposite}{hypotenuse}}$

$sina = \dfrac{2}{ \sqrt{5} }$

$\boxed{\bf cosec=\dfrac{1}{sin}}$

$coseca = \dfrac{ \sqrt{5} }{2}$

$\boxed{\bf cos=\dfrac{adjacent}{hypotenuse}}$

$cosa = \dfrac{1}{ \sqrt{5} }$

$\boxed{\bf sec=\dfrac{1}{cos}}$

$seca = \sqrt{5}$

$\boxed{\bf cot=\dfrac{1}{tan}}$

$cota = \dfrac{1}{2}$

Answer 2

Given,

$\tan( \theta) = 2$

To find out,

The other trigonometric ratios, i.e sin,cos,cosec,sec,cot.

Solution:

$\huge{\tan( \theta) = \frac{opposite \: to \: \theta}{adjacent \: to \: \theta} } = \frac{2}{1}$

In the figure,

$opposite \: side \: to \: \theta = bc = 2$

$adjacent \: side \: to \: \theta \: = ab = 1$

Now,we have to find out the hypotenuse.

By pythagoras theorem: In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the length of the other two sides.

${ac}^{2} = {ab}^{2} + {bc}^{2}$

${ac}^{2} = {1}^{2} + {2}^{2}$

${ac}^{2} = 1 + 4$

${ac}^{2} = 5$

$ac = \sqrt{5}$

Trigonometric ratios:

$\huge \sin( \theta) = \frac{opposite \: side \: to \: \theta}{hypotenuse} = \frac{bc}{ac} = \frac{2}{ \sqrt{5} }$

$\huge\cos( \theta) = \frac{adjacent \: side \: to \: \theta \: }{hypotenuse} = \frac{ab}{ac} = \frac{1}{ \sqrt{5} }$

$\huge \cosec(theta) = \frac{hypotenuse}{opposite \: side \: to \: \theta} = \frac{ \sqrt{5} }{2}$

$\huge \sec( \theta) = \frac{hypotenuse}{adjacent \: side \: to \: \theta} = \frac{ \sqrt{5} }{1}$

$\huge \cot( \theta) = \frac{adjacent \: side \: to \theta}{opposite \: side \: to \theta} = \frac{1}{2}$