Question

can anyone solve it?​

Answer 1

Step-by-step explanation:

Answer 2

Solution

To evaluate: $\sqrt{\dfrac{1+\cos\alpha}{1-\cos\alpha} } +\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha} }$

Let's make the denominator $1-\cos^2\alpha=\sin^2\alpha$, by multiplying $1+\cos\alpha$ and $1-\cos\alpha$ respectively.

$=\sqrt{\dfrac{(1+\cos\alpha)^2}{1-\cos^2\alpha} } +\sqrt{\dfrac{(1-\cos\alpha)^2}{1-\cos^2\alpha} }$

$=\sqrt{\dfrac{(1+\cos\alpha)^2}{\sin^2\alpha} } +\sqrt{\dfrac{(1-\cos\alpha)^2}{\sin^2\alpha} }$

In the first quadrant, or for acute angle, it does not matter to say that

$=\dfrac{1+\cos\alpha}{\sin\alpha} +\dfrac{1-\cos\alpha}{\sin\alpha}=\boxed{2\csc\alpha}$

We are done.

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Graphing

So, what will be the actual graph for obtuse angle, or other?

Let us replace square root with mode.

$=|\dfrac{1+\cos \alpha}{\sin \alpha} |+|\dfrac{1-\cos \alpha}{\sin \alpha}|$

We should be careful, that the mode function has two values that depend on the magnitude of $\alpha$.

[1st Quadrant]

We observe that

• $0<\sin\alpha<1$
• $0<\cos\alpha<1$

Given value

$=\dfrac{1+\cos\alpha}{\sin\alpha} +\dfrac{1-\cos\alpha}{\sin\alpha}$

$=\dfrac{2}{\sin\alpha} =\boxed{2\csc\alpha}$

[2nd Quadrant]

• $0<\sin\alpha<1$
• $-1<\cos\alpha<0$

Given value

$=\dfrac{1+\cos\alpha}{\sin\alpha} +\dfrac{1-\cos\alpha}{\sin\alpha}$

$=\dfrac{2}{\sin\alpha} =\boxed{2\csc\alpha}$

[3rd Quadrant]

• $-1<\sin\alpha<0$
• $-1<\cos\alpha<0$

Given value

$=-\dfrac{1+\cos\alpha}{\sin\alpha} -\dfrac{1-\cos\alpha}{\sin\alpha}$

$=-\dfrac{2}{\sin\alpha} =\boxed{-2\csc\alpha}$

[4th Quadrant]

• $-1<\sin\alpha<0$
• $0<\cos\alpha<1$

Given value

$=-\dfrac{1+\cos\alpha}{\sin\alpha} -\dfrac{1-\cos\alpha}{\sin\alpha}$

$=-\dfrac{2}{\sin\alpha} =\boxed{-2\csc\alpha}$

Now,

$\sqrt{\dfrac{1+\cos \alpha}{1-\cos \alpha} } +\sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha} }=\begin{cases} & 2\csc \alpha\ \text{(In the 1st, 2nd quadrant)} \\ & -2\csc \alpha\ \text{(In the 3rd, 4th quadrant)}\end{cases}$

Which is $2|\csc\alpha|$.

So, small note: $2|\csc\alpha|$ is actually correct!

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The graph is in the attachment. I hope you are satisfied with the answer!