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

If 3π4<α<π\dfrac{{3\pi }}{4} < \alpha < \pi , then find 2cot⁡α+1sin⁡2α\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} ​.

Explanation

Solution

Now in order to find 2cot⁡α+1sin⁡2α\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=1csc⁡x 1+cot⁡2x=csc⁡2x  \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
3π4<α<π.........................................(i)\dfrac{{3\pi }}{4} < \alpha < \pi .........................................\left( i \right)
Now from this given information we have to find 2cot⁡α+1sin⁡2α\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} .
So let’s take the value and simplify it:
2cot⁡α+1sin⁡2α.................................(ii)\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=1csc⁡x 1+cot⁡2x=csc⁡2x  \sin x = \dfrac{1}{{\csc x}} \\\ 1 + {\cot ^2}x = {\csc ^2}x \\\
On substituting it to (ii) we can write:
2cot⁡α+1sin⁡2α=2cot⁡α+(csc⁡2a)\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:

2cot⁡α+1sin⁡2α=2cot⁡α+(csc⁡2α) =2cot⁡α+1+cot⁡2α =cot⁡2α+2cot⁡α+1..........................(iii)  \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:
(a+b)2=a2+2ab+c2{\left( {a + b} \right)^2} = {a^2} + 2ab + {c^2}
Such that we can write (iii) as:

2cot⁡α+1sin⁡2α=cot⁡2α+2cot⁡α+1 =(cot⁡α+1)2 =cot⁡α+1 =1+cot⁡α.................................(iv)  \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 2cot⁡α+1sin⁡2α\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} we get1+cot⁡α1 + \cot \alpha .
Now from (i) we have:
3π4<α<π\dfrac{{3\pi }}{4} < \alpha < \pi
It implies α\alpha is in the II quadrant. Such that there cot⁡α\cot \alpha would be negative.
Therefore we can write on solving 2cot⁡α+1sin⁡2α\sqrt {2\cot \alpha + \dfrac{1}{{{{\sin }^2}\alpha }}} we get1−cot⁡α1 - \cot \alpha .

Note:
General things to be known for solving this question.
Quadrant I: 0  −  π20\; - \;\dfrac{\pi }{2} -> All values are positive.
Quadrant II: π2  −  π\dfrac{\pi }{2}\; - \;\pi -> Only Sine and Cosec values are positive.
Quadrant III: π  −  3π2\pi \; - \;\dfrac{{3\pi }}{2} -> Only Tan and Cot values are positive.
Quadrant IV: 3π2  −  2π\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.