Question

(Mackenzie-SP) Se sen x=1/2 em que 0° < x < 90°, então o valor da expressão y=cos x/tg x + sec x é:

Answer 1

sen x=1/2

cos² x= 1 - sen²x
cos²x= 1-1/4
cos²x= 4-1/4
cos²x=3/4
cos x=√3/2

tg x= sen x/ cos x
tg x= 1/2/√3/2
tg x= 2/2.√3 (racionalização)
tg x= √3/3

sec x= 1/cos x
sec x= 2/√3 (racionalização)
sec x= 2√3/3

y=cos x/tg x+ sec x
y= √3/2/√3/3+2√3/3
y= 3/2+2√3/3
y=9+4√3/6

Answer 2

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$\sf sen(x)=\dfrac{1}{2}\implies sen^2(x)=\bigg(\dfrac{1}{2}\bigg)^2=\dfrac{1}{4}\\\sf cos^2(x)=\dfrac{4-1}{4}=\dfrac{3}{4}\\\sf cos(x)=\dfrac{\sqrt{3}}{\sqrt{4}}=\dfrac{\sqrt{3}}{2}\implies sec(x)=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3} \\\sf tg(x)=\dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\dfrac{\sqrt{3}}{3}$

$\sf y=\dfrac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{3}+\frac{2\sqrt{3}}{3}}\\\sf y=\dfrac{\frac{\sqrt{3}}{2}}{\frac{\diagdown\!\!\!3\sqrt{3}}{\diagdown\!\!\!3}}\\\sf y=\dfrac{\frac{\diagdown\!\!\!\!\!\!\sqrt{3}}{2}}{\diagdown\!\!\!\!\!\!\sqrt{3}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf y=\dfrac{1}{2}}}}}\blue{\checkmark}$