Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De breuk cos p − cos q cos p + cos q is gelijk aan	A. $\boldsymbol{-1 }$
	B. $\large\boldsymbol{-\tan\frac{p^2\,-\,q^2}{4} }$
	C. $\large\boldsymbol{-\cot\frac{p\,+\,q}{2}.\cot\frac{p\,-\,q}{2} }$
	D. $\large\boldsymbol{\tan\frac{p\,+\,q}{2}.\tan\frac{p\,-\,q}{2} }$
	E. $\large\boldsymbol{\tan\frac{p\,+\,q}{2}.\tan\frac{q\,-\,p}{2}}$

[ 5-1515 - op net sinds 18.10.11-(E)-21.7.2024 ]

Translation in E N G L I S H

The fraction cos p − cos q cos p + cos q is equal to	A. $\boldsymbol{-1 }$
	B. $\large\boldsymbol{-\tan\frac{p^2\,-\,q^2}{4} }$
	C. $\large\boldsymbol{-\cot\frac{p\,+\,q}{2}.\cot\frac{p\,-\,q}{2} }$
	D. $\large\boldsymbol{\tan\frac{p\,+\,q}{2}.\tan\frac{p\,-\,q}{2} }$
	E. $\large\boldsymbol{\tan\frac{p\,+\,q}{2}.\tan\frac{q\,-\,p}{2}}$

Oplossing - Solution

$\frac{\cos p - \cos q}{\cos p + \cos q} =\frac{-2.\sin\frac{p+q}{2} \sin\frac{p-q}{2}}{2.\cos\frac{p+q}{2} \cos \frac{p-q}{2}} =- \tan \frac{p+q}{2}.\tan \frac{p-q}{2} =\tan \frac{p+q}{2}.\tan \frac{q-p}{2}$

De breuk cos p − cos q cos p + cos q is gelijk aan	A. \(\boldsymbol{-1 }\)
	B. \(\large\boldsymbol{-\tan\frac{p^2\,-\,q^2}{4} }\)
	C. \(\large\boldsymbol{-\cot\frac{p\,+\,q}{2}.\cot\frac{p\,-\,q}{2} }\)
	D. \(\large\boldsymbol{\tan\frac{p\,+\,q}{2}.\tan\frac{p\,-\,q}{2} }\)
	E. \(\large\boldsymbol{\tan\frac{p\,+\,q}{2}.\tan\frac{q\,-\,p}{2}}\)

The fraction cos p − cos q cos p + cos q is equal to	A. \(\boldsymbol{-1 }\)
	B. \(\large\boldsymbol{-\tan\frac{p^2\,-\,q^2}{4} }\)
	C. \(\large\boldsymbol{-\cot\frac{p\,+\,q}{2}.\cot\frac{p\,-\,q}{2} }\)
	D. \(\large\boldsymbol{\tan\frac{p\,+\,q}{2}.\tan\frac{p\,-\,q}{2} }\)
	E. \(\large\boldsymbol{\tan\frac{p\,+\,q}{2}.\tan\frac{q\,-\,p}{2}}\)