Question

If \[0\le x\le \pi \] and \[\cos x+\sin x=\dfrac{1}{2}\], then \[\tan x\] is equal to
(a) \[\dfrac{\left( 4-\sqrt{7} \right)}{3}\]
(b) \[\dfrac{-\left( 4+\sqrt{7} \right)}{3}\]
(c) \[\dfrac{\left( 1+\sqrt{7} \right)}{4}\]
(d) \[\dfrac{\left( 1-\sqrt{7} \right)}{4}\]

Answer 1

Hint: For solving this question you should know about the trigonometric formulas and general solutions for these. In this problem we have given the equation in sin and cos form so we will change it in the \[\tan x\] form and then we will solve it and will determine the values of d and find the values of \[\tan x\].

Complete step by step answer:
According to our question, it is asked to find the value of \[\tan x\] if \[0\le x\le \pi \] and \[\cos x+\sin x=\dfrac{1}{2}\].
As we can see the equation is given as \[\cos x+\sin x=\dfrac{1}{2}\]
which is in the form of cos and sin and we have to ask for \[\tan x\]. So, we will divide the equation by \[\cos x\] on both sides.
\[\Rightarrow \dfrac{\cos x}{\cos x}+\dfrac{\sin x}{\cos x}=\dfrac{1}{2\cos x}\]
\[\Rightarrow 1+\tan x=\dfrac{\sec x}{2}\]
If we square it both sides, then
\[\Rightarrow {{\left( 1+\tan x \right)}^{2}}={{\left( \dfrac{\sec x}{2} \right)}^{2}}\]
\[\Rightarrow 1+{{\tan }^{2}}x+2\tan x=\dfrac{{{\sec }^{2}}x}{4}\]
\[\Rightarrow 4\left( 1+{{\tan }^{2}}x+2\tan x \right)={{\sec }^{2}}x=1+{{\tan }^{2}}x\]
And if we solve these then we get,
\[\Rightarrow 3{{\tan }^{2}}x+8\tan x+3=0\]
\[\begin{align}
  & D={{d}^{2}}={{b}^{2}}-4ac \\
 & D={{\left( 8 \right)}^{2}}-4\left( 3 \right)\left( 3 \right) \\
 & {{d}^{2}}=64-36=28 \\
 & {{d}^{2}}=28 \\
 & d=\pm 2\sqrt{7} \\
\end{align}\]
So, the two real roots
\[\tan x=\dfrac{-b}{2a}\pm \dfrac{d}{2a}=\dfrac{-8}{6}\pm \dfrac{2\sqrt{7}}{6}\]
           \[=\dfrac{-4\pm \sqrt{7}}{3}\]

So, the correct answer is “Option b”.

Note: While solving these types of questions you have to be careful about the formulas which we are using to convert one form to another form. Because if we use the formula which also gives answers but that has a very lengthy process, there are also chances to make calculation mistakes also.