Question

How do you find the value of six trigonometric functions given $\cot \theta = - 3$ and $\cos \theta > 0$ .

Answer 1

Hint : Here we have to find all the six trigonometric function values such that one of the functions is given to us. We will use some standard trigonometric identities to get the values of remaining functions. Also we are given that cot function is negat

** Complete step-by-step answer** :
Given that $\cot \theta = - 3$
We know that, ${\sin ^2}\theta = \dfrac{1}{{1 + {{\cot }^2}\theta }}$
Putting the value of $\cot \theta = - 3$ we get,
${\sin ^2}\theta = \dfrac{1}{{1 + {{\left( { - 3} \right)}^2}}}$
On taking the square and adding we get,
${\sin ^2}\theta = \dfrac{1}{{10}}$
Taking roots on both the sides we get,
$\sin \theta = \pm \dfrac{1}{{\sqrt {10} }}$
But since cot function is negative and cos function is positive the angle is in the fourth quadrant. So the sine function will be negative.
Thus $\sin \theta = - \dfrac{1}{{\sqrt {10} }}$
This is the first value. Now let’s find the remaining one by one.
We know that ${\cos ^2}\theta = 1 - {\sin ^2}\theta$
Thus ${\cos ^2}\theta = 1 - {\left( {\dfrac{1}{{\sqrt {10} }}} \right)^2}$
Performing the square we get,
${\cos ^2}\theta = 1 - \dfrac{1}{{10}}$
Taking LCM and solving further we get,
${\cos ^2}\theta = \dfrac{9}{{10}}$
Taking roots on both the sides we get,
$\cos \theta = \pm \sqrt {\dfrac{9}{{10}}}$
Since 9 is the perfect square of 3 we can write 3 in numerator,
$\cos \theta = \pm \dfrac{3}{{\sqrt {10} }}$
But since cos is given positive we will take the plus sign.
$\cos \theta = \dfrac{3}{{\sqrt {10} }}$
This is the second function.
Now we have $sec\theta = \dfrac{1}{{\cos \theta }}$
Thus just taking reciprocal of cos function we get,
$sec\theta = \dfrac{{\sqrt {10} }}{3}$
This is the third function.
Now we know that,
$\tan \theta = \dfrac{1}{{\cot \theta }}$
So just putting the given value we get,
$\tan \theta = \dfrac{1}{{ - 3}}$
Taking minus sign in numerator we get,
$\tan \theta = \dfrac{{ - 1}}{3}$
This is the fourth function.
Now we have $\cos ec\theta = \dfrac{1}{{\sin \theta }}$
Just putting the value 0f sine function
$\cos ec\theta = \dfrac{1}{{ - \dfrac{1}{{\sqrt {10} }}}}$
Taking the terms in proper order we get,
$\cos ec\theta = - \sqrt {10}$
This is the last function.
Let’s tabulate the values.

$\sin \theta = - \dfrac{1}{{\sqrt {10} }}$	4. $\cos ec\theta = - \sqrt {10}$
$\cos \theta = \dfrac{3}{{\sqrt {10} }}$	5. $sec\theta = \dfrac{{\sqrt {10} }}{3}$
$\tan \theta = \dfrac{{ - 1}}{3}$	6. $\cot \theta = - 3$

Note : Here the thing you should note is when we get two values of a function we need to take the value according to the quadrant or if any other given data such as where the angle lies. Also note that the only thing we need to find is any one or two values because rest other functions can be obtained using the reciprocal or trigonometric identity.