Question

If $\sin \theta = \dfrac{{\sqrt 3 }}{2}$, find the values of all T-ratios of $\theta$.

Answer 1

In this question, we are given the value of $\sin \theta$ and we are asked to find all other trigonometric ratios. We have to find all the trigonometric ratios, step by step. We know the identity, ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Using this relation $\cos \theta$ can be obtained. Now, we have the value of $\sin \theta$ and $\cos \theta$. By these values, $\tan \theta$ can be calculated. Now, other remaining trigonometric ratios can be calculated by finding the reciprocal of these trigonometric ratios.

Complete step-by-step solution:
Now, according to the question, it is given that
$\sin \theta = \dfrac{{\sqrt 3 }}{2}$....................….. (1)
As we know the identity,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
Taking ${\sin ^2}\theta$ to the right side, we get
$\Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta$
Now, $\cos \theta$ can be expressed in terms of $\sin \theta$.
Taking square root on both sides, we get,
$\Rightarrow \cos \theta = \sqrt {1 - {{\sin }^2}\theta }$
Substitute the value of $\sin \theta$ from the equation (1),
$\Rightarrow \cos \theta = \sqrt {1 - {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}$
Square the term inside the square root,
$\Rightarrow \cos \theta = \sqrt {1 - \dfrac{3}{4}}$
Take LCM inside the square root,
$\Rightarrow \cos \theta = \sqrt {\dfrac{{4 - 3}}{4}}$
Subtract the values in the numerator,
$\Rightarrow \cos \theta = \sqrt {\dfrac{1}{4}}$
Simplify the term,
$\Rightarrow \cos \theta = \dfrac{1}{2}$......................….. (2)
As we know,
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Substitute the values from equation (1) and (2),
$\Rightarrow \tan \theta = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}$
Cancel out the common factors,
$\Rightarrow \tan \theta = \sqrt 3$....................….. (3)
We have to find other remaining trigonometric ratios that are $\operatorname{cosec} \theta ,\sec \theta$ and $\cot \theta$.
As we know,
$\operatorname{cosec} \theta = \dfrac{1}{{\sin \theta }}$
Substitute the value from equation (1),
$\Rightarrow \operatorname{cosec} \theta = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}$
Simplify the term,
$\Rightarrow \operatorname{cosec} \theta = \dfrac{2}{{\sqrt 3 }}$
As we know,
$sec\theta = \dfrac{1}{{\cos \theta }}$
Substitute the value from equation (2),
$\Rightarrow sec\theta = \dfrac{1}{{\dfrac{1}{2}}}$
Simplify the term,
$\Rightarrow \sec \theta = 2$
As we know,
$\cot \theta = \dfrac{1}{{\tan \theta }}$
Substitute the value from equation (3),
$\Rightarrow \cot \theta = \dfrac{1}{{\sqrt 3 }}$

Note: This question can also be solved by using the Pythagoras theorem.
We have,
$\Rightarrow \sin \theta = \dfrac{{\sqrt 3 }}{2}$.................….. (1)
As we know,
$\sin \theta = \dfrac{{{\text{height}}}}{{{\text{hypotenuse}}}}$
So, the value of height and hypotenuse is,
$\Rightarrow$ Height $= \sqrt 3$
$\Rightarrow$ Hypotenuse $= 2$
Using Pythagoras theorem, we can find the base.
$\Rightarrow {\text{base}} = \sqrt {{{\left( {{\text{hypotenuse}}} \right)}^2} - {{\left( {{\text{height}}} \right)}^2}}$
Substitute the values,
$\Rightarrow$ base $= \sqrt {{2^2} - {{\left( {\sqrt 3 } \right)}^2}}$
Square the terms,
$\Rightarrow$ base $= \sqrt {4 - 3}$
Subtract the values,
$\Rightarrow$ base $= \sqrt 1$
Simplify the terms,
$\Rightarrow$ base $= 1$
As we know,
$\cos \theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}$
Substitute the values,
$\Rightarrow \cos \theta = \dfrac{1}{2}$....................….. (2)
As we know,
$\tan \theta = \dfrac{{{\text{height}}}}{{{\text{base}}}}$
Substitute the values,
$\Rightarrow \tan \theta = \dfrac{{\sqrt 3 }}{1}$
Simplify the terms,
$\Rightarrow \tan \theta = \sqrt 3$..............….. (3)
We have to find other remaining trigonometric ratios that are $\operatorname{cosec} \theta ,\sec \theta$ and $\cot \theta$.
As we know,
$\operatorname{cosec} \theta = \dfrac{1}{{\sin \theta }}$
Substitute the value from equation (1),
$\Rightarrow \operatorname{cosec} \theta = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}$
Simplify the term,
$\Rightarrow \operatorname{cosec} \theta = \dfrac{2}{{\sqrt 3 }}$
As we know,
$sec\theta = \dfrac{1}{{\cos \theta }}$
Substitute the value from equation (2),
$\Rightarrow sec\theta = \dfrac{1}{{\dfrac{1}{2}}}$
Simplify the term,
$\Rightarrow \sec \theta = 2$
As we know,
$\cot \theta = \dfrac{1}{{\tan \theta }}$
Substitute the value from equation (3),
$\Rightarrow \cot \theta = \dfrac{1}{{\sqrt 3 }}$