3 years ago

Brainliest for answer

Mathematics

1 answer:

Arlecino [84]3 years ago

3 0

4/15 !! hope it helps loll

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$\huge\boxed{106}$

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GuDViN [60]

Recall that

$\cot(u+v)=\dfrac{\cos(u+v)}{\sin(u+v)}=\dfrac{\cos u\cos v-\sin u\sin v}{\sin u\cos v+\cos u\sin v}=\dfrac{\cot u\cot v-1}{\cot v+\cot u}$

Also, recall that for all $\theta$ ,

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With $\frac{3\pi}2$ , we can expect $\cos u>0$ , and with $0$ , $\sin v>0$ . So from the above identity it follows that

$\cos u=\sqrt{1-\sin^2u}=\dfrac45\implies\cot u=\dfrac{\cos u}{\sin u}=-\dfrac43$

and

$\sin v=\sqrt{1-\cos^2v}=\dfrac8{17}\implies\cot v=\dfrac{\cos v}{\sin v}=\dfrac{15}8$

and so

$\cot(u+v)=\dfrac{-\frac43\frac{15}8-1}{\frac{15}8-\frac43}=\boxed{-\dfrac{84}{13}}$

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Step-by-step explanation:

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