Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\!\!\!\sin^2\alpha\\+\cos^2\alpha\\+\tan^2\alpha$ is gelijk aan	A. \(\large\boldsymbol{\frac {1} {\sin^2\alpha} }\)
	B. \(\large\boldsymbol{\frac {1} {\cos^2\alpha} }\)
	C. \(\large\boldsymbol{\frac {1\,+\:\sin^2\alpha} {\cos^2\alpha} }\)
	D. \(\boldsymbol{ {\tan^2 \alpha} }\)
	E. \(\large\boldsymbol{\frac {1} {\tan^2\alpha} }\)

[ 4-2029 - op net sinds 7.3.16-(E)-21.7.2024 ]

Translation in E N G L I S H

$https://latex.codecogs.com/png.image?\dpi{110}\!\!\!\sin^2\alpha\\+\cos^2\alpha\\+\tan^2\alpha$ is equal to	A. \(\large\boldsymbol{\frac {1} {\sin^2\alpha} }\)
	B. \(\large\boldsymbol{\frac {1} {\cos^2\alpha} }\)
	C. \(\large\boldsymbol{\frac {1\,+\:\sin^2\alpha} {\cos^2\alpha} }\)
	D. \(\boldsymbol{ {\tan^2 \alpha} }\)
	E. \(\large\boldsymbol{\frac {1} {\tan^2\alpha} }\)

Oplossing - Solution

Samensmelting van twee basisformules van de goniometrie :
$\sin ^2\alpha +\cos^2 \alpha =2 \quad en \quad 1+\tan ^2 \alpha =\frac{1}{\cos ^2\alpha }$

gricha