Question

If I tan theta is equal to 4 show that 5 sin theta minus 3 cos theta divided by 5 sin theta + 2 cos theta is equal to 1 by 6

Answer 1

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According to the question,

$\tan( \theta) = 4$

We know that,

$\tan( \theta) = \frac{perpendicular}{base \: } \\ \\ \tan( \theta) = \frac{4}{1} = \frac{perpendicular}{base \: } \:$

Using Pythagoras theorem,

$\small\implies \: {(hypt.)}^{2} = {(perpend.) }^{2} + {(base)}^{2} \\ \\ \implies \: {(hypt.)}^{2} =16 + 1 \\ \\ \implies \: \boxed{ hypt. = \sqrt{17} } \\ \\ hypt. = hypotenous \\ \\ perpend. = perpendicular$

To Prove :-

$\frac{5 \sin( \theta) - 3 \cos( \theta) }{5 \sin( \theta) + 2 \cos( \theta) } = \frac{17}{22}$

Proof :-

$\because \: hypt. = \sqrt{17} \: \\ \: \: \: \: \: perpend. = 4 \\ \: \: \: \: \: base = 1 \\ then$

We know that,

$\sin( \theta) = \frac{perpend.}{hypt.} = \frac{4}{ \sqrt{17} } \\ \\ \cos( \theta) = \frac{base}{hypt.} = \frac{1}{ \sqrt{17} }$

Now taking left hand side of ,

$\implies \: \frac{5 \sin( \theta) - 3 \cos( \theta) }{5 \sin( \theta) + 2 \cos( \theta) }$

Substitute the above values,

$\implies \: \frac{5 \times \frac{4}{ \sqrt{17} } - 3 \times \frac{1}{ \sqrt{17} } }{5 \times \frac{4}{ \sqrt{17} } + 2 \times \frac{1}{ \sqrt{17} } } \\ \\ \implies \: \frac{ \frac{20 - 3}{ \sqrt{17} } }{ \frac{20 + 2}{ \sqrt{17} } } \\ \\ \implies \: \frac{17}{22}$

Hence proved.

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