Math, asked by monjyotiboro, 19 days ago

can anyone solve it?​

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Answers

Answered by mahek339575
0

Step-by-step explanation:

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Answered by user0888
8

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!

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