Question

The solution of $4{\sin ^2}x + {\tan ^2}x + {\operatorname{cosec} ^2}x + {\cot ^2}x - 6 = 0$ is
A) $n\pi \pm \dfrac{\pi }{4}$
B) $2n\pi \pm \dfrac{\pi }{4}$
C) $n\pi + \dfrac{\pi }{3}$
D) $n\pi - \dfrac{\pi }{6}$

Answer 1

Hint: First try to make the formula in terms of ${\left( {a - b} \right)^2}$. After that solve the term such that it will get the form $\sin \alpha = \sin \beta $ or $\tan \alpha = \tan \beta $ and then apply the formula $\alpha = n\pi \pm \beta $ to find the solution of the equation.

Complete step-by-step answer:
Given: - $4{\sin ^2}x + {\tan ^2}x + {\operatorname{cosec} ^2}x + {\cot ^2}x - 6 = 0$
It can be written as,
$ \Rightarrow {\left( {2\sin x} \right)^2} + {\left( {\operatorname{cosec} x} \right)^2} - 4 + {\tan ^2}x + {\cot ^2}x - 2 = 0$
Solving it further, we get the equation as,
$ \Rightarrow {\left( {2\sin x} \right)^2} + {\left( {\operatorname{cosec} x} \right)^2} - 2\left( {2\sin x} \right)\left( {\operatorname{cosec} x} \right) + {\tan ^2}x + {\cot ^2}x - 2\left( {\tan x} \right)\left( {\cot x} \right) = 0$...........….. (1)
As we know the formula,
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
Putting the values of equation (1) in the obtained equation we get the new equation as:
$ \Rightarrow {\left( {2\sin x - \operatorname{cosec} x} \right)^2} + {\left( {\tan x - \cot x} \right)^2} = 0$
Since the sum of the square is zero then each term should be zero. Then,
$ \Rightarrow {\left( {2\sin x - \operatorname{cosec} x} \right)^2} = 0,{\left( {\tan x - \cot x} \right)^2} = 0$
Take the square root on both sides,
$ \Rightarrow \left( {2\sin x - \operatorname{cosec} x} \right) = 0$...........…. (2)
$ \Rightarrow \left( {\tan x - \cot x} \right) = 0$............…. (3)
From equation (2), move $\operatorname{cosec} x$ on the right side,
$ \Rightarrow 2\sin x = \operatorname{cosec} x$
Replace $\operatorname{cosec} x$ with $\dfrac{1}{{\sin x}}$,
$ \Rightarrow 2\sin x = \dfrac{1}{{\sin x}}$
Cross-multiply the terms,
$ \Rightarrow 2{\sin ^2}x = 1$
Divide both sides by 2,
$ \Rightarrow {\sin ^2}x = \dfrac{1}{2}$
Take the square root on both sides,
$ \Rightarrow \sin x = \pm \dfrac{1}{{\sqrt 2 }}$
Substitute the value of $\dfrac{1}{{\sqrt 2 }}$ in terms of sine.
$ \Rightarrow \sin x = \pm \sin \dfrac{\pi }{4}$
The solution of $\left( {2\sin x - \operatorname{cosec} x} \right) = 0$ is,
$\therefore x = n\pi \pm \dfrac{\pi }{4}$
From equation (2), move $\cot x$ on the right side,
$ \Rightarrow \tan x = \cot x$
Replace $\cot x$ with $\dfrac{1}{{\tan x}}$,
$ \Rightarrow \tan x = \dfrac{1}{{\tan x}}$
Cross-multiply the terms,
$ \Rightarrow {\tan ^2}x = 1$
Take the square root on both sides,
$ \Rightarrow \tan x = \pm 1$
Substitute the value of 1 in terms of tan.
$ \Rightarrow \tan x = \pm \tan \dfrac{\pi }{4}$
The solution of $\left( {\tan x - \cot x} \right) = 0$ is,
$\therefore x = n\pi \pm \dfrac{\pi }{4}$

Hence, option (B) is the correct answer.

Note: Whenever you are stuck with these types of problems you should always think about which identity, we can use so that we can prove what has been asked like here we have to make the given equation such that we can use the formula of ${\left( {a - b} \right)^2}$. Proceeding like this will make your solution correct.
One should know the formulas of trigonometry to simplify the given expression. Also, one must know the general solution when two trigonometric expressions are given equal.