Question

If sec 0 = 13/5
show that
2 sin theta - 3 cos theta / 4 sin theta - 9 cos theta = 3​

Answer 1

Answer:

your answer attached in the photo

Answer 2

Answer:

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$sec \: θ = \frac{13}{5}$

$cos \: θ = \frac{1}{secθ}$

$\cosθ = \frac{5}{13}$

Now

$\sin θ = \sqrt{1 - ( \frac{5}{13}) {}^{2} }$

$\sinθ = \sqrt{ \frac{144}{169} }$

$sin \: θ = \frac{12}{13}$

So

$\frac{2 \: sin \: θ - 3 \: cos \:θ }{4 \: sin \: θ - 9 \: cos \: θ}$

$→ \frac{2 \times \frac{12}{13} - 3 \times \frac{5}{13} }{4 \times \frac{12}{13} - 9 \times \frac{5}{13} }$

$→ \frac{9}{3}$

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Hence proved