Question

sec theta= 13/12 find sin theta- 2 cos theta/tan theta - cot theta
plss help​

Answer 1

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Answer 2

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$if sec Θ = \dfrac{13}{12}$

$find ;\dfrac {sin Θ-2 cosΘ}{tan Θ - cot Θ}$

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$if sec Θ = \dfrac{13}{12}$

$if sec Θ = \dfrac{hypotenuse}{base}$

• applying Pythagoras theoram

$h^2 = p^2 - b^2$

$13^2 = p^2 - 12^2$

$169- 144 = p^2$

$25 = p^2$

$p= \sqrt25$

$p= 5$

• finding value of sinΘ, cosΘ, tanΘ, cotΘ

$sinΘ = \dfrac{p}{h} = \dfrac{5}{13}$

$cosΘ= \dfrac{b}{h} = \dfrac{12}{13}$

$tan Θ = \dfrac{p}{b} = \dfrac{5}{12}$

$cotΘ = \dfrac{b}{p} = \dfrac{12}{5}$

• putting value in equation

$\dfrac {sin Θ-2 cosΘ}{tan Θ - cot Θ}$

$\dfrac{\dfrac{12}{13}-\dfrac{2*12}{13}}{\dfrac{2*5}{12}-\dfrac{12}{5}}$

$\dfrac{\dfrac{12}{13}-\dfrac{24}{13}}{\dfrac{10}{12}-\dfrac{12}{5}}$

$\dfrac{\dfrac{12-24}{13}}{\dfrac{50-120}{60}}$

$\dfrac{\dfrac{-12}{13}}{\dfrac{-70}{60}}$

$\dfrac{-12*60}{13*-70}$

$\dfrac{12*6}{13*7}$

$\dfrac{72}{91}$

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