Question

5cot = 12, find the value of cosec + sec​

Answer 1

Step-by-step explanation:

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Answer 2

$5 \cot θ = 12$

$cot θ = \frac{12}{5}$

$\cot θ = \frac{b}{p} = \frac{12}{5}$

let base = 12k and perpendicular= 5k

$\blue{by \: pythagoras \: theorm}$

$h {}^{2} = p {}^{2} + b {}^{2}$

${( h)}^{2} = {(5k)}^{2} + (12k) {}^{2}$

${h}^{2} = 25k {}^{2} + 144k {}^{2}$

$h {}^{2} = 169k {}^{2}$

$h = \sqrt{169k {}^{2} }$

$h = 13k$

$\cosec θ = \frac{h}{p} = \frac{13k}{5k} = \frac{13}{5}$

$\sec θ = \frac{h}{b} = \frac{13k}{12k} = \frac{13}{12}$

$cosec θ + sec θ = \frac{13}{5} + \frac{13}{12}$

$= > \frac{156 +65 }{60}$

$= > \green {\frac{221}{60} }$

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