Question

If $\sin \theta + \cos \theta = \sqrt 2$ , then find the value of $\tan \theta + \cot \theta$ .

Answer 1

Here we are given one value and we need to find the value of another trigonometric function. So we need to simplify the above term which we need to find and use the given value in it to solve the above problem. Here we are given the value of $\sin \theta + \cos \theta = \sqrt 2$ so after squaring both sides we will get the value of $\sin 2\theta$ and afterwards we can simplify $\tan \theta + \cot \theta$ and get the value required.

Complete step by step solution:
Here we are given that $\sin \theta + \cos \theta = \sqrt 2$ and we need to find the value of $\tan \theta + \cot \theta$
So first of all we can simplify the equation given as:
$\sin \theta + \cos \theta = \sqrt 2$ $- - - - - (1)$
Squaring both the sides of the equation (1), we get:
${(\sin \theta + \cos \theta )^2} = {(\sqrt 2 )^2}$
We know that ${(a + b)^2} = {a^2} + {b^2} + 2ab$ so we can apply this formula in the above equation and get:
${\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = 2$ $- - - - - (2)$
From the trigonometric properties we also know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$
So substituting it in the above equation (2) we get:
$1 + 2\sin \theta \cos \theta = 2$
$2\sin \theta \cos \theta = 1$ $- - - - - (3)$
Now we need to simplify the term we need to find:
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta {\text{ }}}}{\text{ and cot}}\theta = \dfrac{{\cos \theta }}{{\sin \theta {\text{ }}}}$
So substituting these values in $\tan \theta + \cot \theta$
$\tan \theta + \cot \theta$ $= \dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}$
So we can take the LCM on the right hand side and we know that the LCM of both the denominators which are $\sin \theta ,\cos \theta = \sin \theta .\cos \theta$
Now taking the LCM we get that:
$\dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }} = \dfrac{{\sin \theta (\sin \theta ) + \cos \theta (\cos \theta )}}{{\sin \theta \cos \theta }}$
$= \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta \cos \theta }}$ $$
Again putting the value of ${\sin ^2}\theta + {\cos ^2}\theta = 1$ in the above equation we get:
$\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta \cos \theta }} = \dfrac{1}{{\sin \theta \cos \theta }}$ $- - - - (5)$
Form the equation $(3)$ we have got the value of $2\sin \theta \cos \theta = 1$
So we can rewrite it as $\sin \theta \cos \theta = \dfrac{1}{2}$
Now substituting the above value of $\sin \theta \cos \theta = \dfrac{1}{2}$ in equation (5), we get:
$\tan \theta + \cot \theta$ $= \dfrac{1}{{\sin \theta \cos \theta }} = \dfrac{1}{{\dfrac{1}{2}}} = 2$
Hence we get the value which is required of $\tan \theta + \cot \theta$ as $2$ .

So $\tan \theta + \cot \theta$ $= 2$

Note:
Here we must know all the trigonometric properties so that we can apply the formula easily and get the value whichever is required. We have the formula like:
${\sin ^2}\theta + {\cos ^2}\theta = 1$
${\tan ^2}\theta + 1 = {\sec ^2}\theta$
$1 + {\cot ^2}\theta = {\text{cose}}{{\text{c}}^2}\theta$
We can also solve this problem by the Hit and Trial method. In $\sin \theta + \cos \theta = \sqrt 2$ if we let $\theta = 45^\circ$
Then this equation is satisfying. So we can put this value in the required equation which is
$\tan \theta + \cot \theta$ $= \tan 45 + \cot 45 = 1 + 1 = 2$
This method can be used for saving the time in the competition.