Question

❤ 50 points❤ please answer with explanation

Answer 1

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=> SOLUTION :

sin@ = -5/13 (given)

we know that :-

sin@ = p/h = -5/13

by using Pythagoras theorem we will find the base of the triangle.

=> h² = b² + p²

=> (AC)² = (BC)² + (AB)²

=> (13)² = (BC)² + (-5)²

=> 169 = (BC)² + 25

=> (BC)²= 169 - 25

=> BC = √144

=> BC = 12

this all are in third quard so, only tan@ and cot@ will in possitive sign .

=> Given Equation

=> 5cot²@ +12tan@ +13cosec@ = ?

=> we know the trigonometric values

=> cot@ = b/p = 12/5

=> tan@ = p/b = 5/12

=> cosec@ = h/p = 13/-5 = -13/5

let's solve this =>

=> 5 × 12/5 + 12× 5/12 + 13× -13/5

=> 12 + 5 - 169/5

=> 60+ 25 - 169/5

=> 85-169/5

=> -84/5 your answer

=> 180°< @ < 270°

=> this is the third quard so,

Answer 2

$\text{given that} \: {180}^{o} < \theta < {270}^{o} \\ \\ \text{so} \: \theta \: \text{ lies in third quadrant}$

In third quadrant, only tan and cot functions are positive. Rest functions are negative.
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We are given sin value. We need the cot, tan and cosec value. So if we find the cos value, we can solve it.

$\sin( \theta) = \frac{ - 5}{13} \\ \\ we \: know: \: { \sin( \theta) }^{2} + { \cos( \theta) }^{2} = 1 \\ \\ \cos( \theta) = \sqrt{1 - { \sin( \theta) }^{2}} \\ \\ \cos( \theta) = \sqrt{1 - { (\frac{5}{13}) }^{2} }\\ \\ \cos( \theta) = \sqrt{1 - \frac{25}{169}} = \sqrt{ \frac{169 - 25}{169} } \\ \\ \cos( \theta) = \sqrt{ \frac{144}{169} }\\ \\ \cos( \theta) = - \frac{12}{25} \\ ( \text{becuse cos is - ve in 3rd quadrant})$
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$\text{Find all required values:}$

$\cot( \theta) = \frac{ \cos( \theta) }{ \sin(\theta) } = \frac{ \frac{ - 12}{13} }{ \frac{ - 5}{13} } = \frac{ - 12}{13} \times \frac{13}{ - 5} = \frac{12}{5}$

$\tan(\theta) = \frac{1}{ \cot( \theta) } = \frac{1}{ \frac{12}{5} } = \frac{5}{12}$

$\cosec( \theta) = \frac{1}{ \sin( \theta) } = \frac{1}{ \frac{ - 5}{13} } = \frac{ - 13}{5}$
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$5 { \cot( \theta) }^{2} + 12 \tan( \theta) + 13 \cosec( \theta) \\ \\ = 5 \times { (\frac{12}{5} )}^{2} + 12 \times \frac{5}{12} + 13 \times ( \frac{ - 13}{5} ) \\ \\ = 5 \times \frac{144}{25} + \: 5 \: - \frac{ 169}{5} \\ \\ = \frac{144}{5} + \: 5 \: - \frac{ 169}{5} \\ \\ = \frac{144 - 169}{5} + 5 \\ \\ = \frac{ - 25}{5} + 5 \\ \\ = - 5 + 5 \\ \\ = \boxed{ \huge{0}}$