Question

1 + cos theta / 1 - cos theta = 16/9 ; 1 + cot theta / 1 - cot theta

Answer 1

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Answer 2

The given question is 1 + cos theta / 1 - cos theta = 16/9 ;

we have to find the value of 1 + cot theta / 1 - cot theta

The given expression is

$\frac{1 + \cos( \: theta) }{1 - \cos(theta) } = \frac{16}{9}$

we have to find the value of

$\frac{1 + \cot(theta) }{1 - \cot(theta) }$

In the expression of the question, add the numerator and denominator in the place of the numerator, and subtract the numerator and the denominator in the place of the denominator.

$\frac{1 \cos(theta) + 1 - \cos(theta) }{1 + \cos(theta) - 1 + \cos(theta) } = \frac{16 + 9}{16 - 9}$

in the numerator, positive and negative cos theta will get cancelled and in the denominator, positive and negative one will get cancelled.

After the cancellation, we get the values as

$\frac{2}{2 \cos(theta) } = \frac{25}{7}$

the digit 2 in the numerator and denominator gets cancelled, and then we have

$\frac{1}{ \cos(theta) } = \frac{25}{7 \\ }$

Therefore, the

$\cos(theta) = \frac{7}{25}$

The formula to determine the cot theta is

$\frac{1}{ \tan(theta) } = \frac{b}{p}$

Here b= 7 and h = 25

$p = \sqrt{ {25}^{2} - {7}^{2} } \\ = \sqrt{625 - 49} = \sqrt{576} = 24$

The value of

$\frac{b}{p} = \frac{7}{24}$

Therefore, the resultant value will be

$\frac{1 + \cot(theta) }{1 - \cot(theta) } = \frac{1 + \frac{7}{24} }{1 - \frac{7}{24} } = \frac{ \frac{31}{24} }{ \frac{17}{24} } = \frac{31}{17}$

The final answer is 31/17

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