Question

If \[\dfrac{{3\pi }}{4} < \alpha < \pi \] , then find \[\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} \] ​.

Answer 1

Hint: Now in order to find \[\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} \] we should work with one side at a time and manipulate it to the other side.
Using some basic trigonometric identities like the below one we can simplify the above expression.
$
  \sin x = \dfrac{1}{{\csc x}} \\
  1 + {\cot ^2}x = {\csc ^2}x \\
 $
In order to verify the given expression we have to use the above identities and express our given expression in that form and thereby find the value.

Complete step by step answer:
Given
\[\dfrac{{3\pi }}{4} < \alpha < \pi .........................................\left( i \right)\]
Now from this given information we have to find \[\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} \].
So let’s take the value and simplify it:
\[\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} .................................\left( {ii} \right)\]
Now for simplifying it we have to take basic trigonometric identities and properties. Such that we know that:
$
  \sin x = \dfrac{1}{{\csc x}} \\
  1 + {\cot ^2}x = {\csc ^2}x \\
 $
On substituting it to (ii) we can write:
\[\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} = \sqrt {2\cot \alpha + \left( {{{\csc }^2}a} \right)} \]
Now substituting the second identity we can write:
\[
  \sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} = \sqrt {2\cot \alpha + \left( {{{\csc }^2}\alpha } \right)} \\
   = \sqrt {2\cot \alpha + 1 + {{\cot }^2}\alpha } \\
   = \sqrt {{{\cot }^2}\alpha + 2\cot \alpha + 1} ..........................\left( {iii} \right) \\
 \]
Now on observing (iii) we can see that it is of the form:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {c^2}$
Such that we can write (iii) as:
\[
  \sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} = \sqrt {{{\cot }^2}\alpha + 2\cot \alpha + 1} \\
   = \sqrt {{{\left( {\cot \alpha + 1} \right)}^2}} \\
   = \cot \alpha + 1 \\
   = 1 + \cot \alpha .................................\left( {iv} \right) \\
 \]
So on solving \[\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} \]we get\[1 + \cot \alpha \].
Now from (i) we have:
\[\dfrac{{3\pi }}{4} < \alpha < \pi \]
It implies \[\alpha \] is in the II quadrant. Such that there \[\cot \alpha \] would be negative.
Therefore we can write on solving \[\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} \] we get\[1 - \cot \alpha \].

Note:
General things to be known for solving this question.
Quadrant I: $0\; - \;\dfrac{\pi }{2}$ -> All values are positive.
Quadrant II: $\dfrac{\pi }{2}\; - \;\pi $ -> Only Sine and Cosec values are positive.
Quadrant III: $\pi \; - \;\dfrac{{3\pi }}{2}$ -> Only Tan and Cot values are positive.
Quadrant IV: $\dfrac{{3\pi }}{2}\; - \;2\pi $ -> Only Cos and Sec values are positive.
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.