Answer: [tex]\frac{2\sqrt{2}-\sqrt{3}}{2}[/tex]
Step-by-step explanation (using the Unit Circle):
csc [tex]\frac{\pi}{4}[/tex] - cos [tex]\frac{\pi}{6}[/tex]
csc = [tex]\frac{1}{sin}[/tex]
sin [tex]\frac{\pi}{4}[/tex] = [tex]\frac{\sqrt{2}}{2}[/tex]
→ csc [tex]\frac{\pi}{4}[/tex] = [tex]\frac{2}{\sqrt{2}}[/tex]= [tex]{\sqrt{2}}[/tex]
cos [tex]\frac{\pi}{6}[/tex] = [tex]\frac{\sqrt{3}}{2}[/tex]
csc [tex]\frac{\pi}{4}[/tex] - cos [tex]\frac{\pi}{6}[/tex]
= [tex]{\sqrt{2}}[/tex] - [tex]\frac{\sqrt{3}}{2}[/tex]
= [tex]{\sqrt{2}}*(\frac{2}{2})[/tex] - [tex]\frac{\sqrt{3}}{2}[/tex]
= [tex]\frac{2\sqrt{2}}{2}[/tex] - [tex]\frac{\sqrt{3}}{2}[/tex]
= [tex]\frac{2\sqrt{2}-\sqrt{3}}{2}[/tex]
Step-by-step explanation (using the special triangles):
[tex]\frac{\pi}{4}[/tex] = 45°
a 45°-45°-90° triangle has sides with proportions of: 1 - 1 - √2
csc = [tex]\frac{hypotenuse}{opposite}[/tex] = [tex]\frac{\sqrt{2}} {1} =\sqrt{2}[/tex]
[tex]\frac{\pi}{6}[/tex] = 30°
a 30°-60°-90° triangle has sides with proportions of: 1 - √3 - 2
cos = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{\sqrt{3}}{2}[/tex]
csc [tex]\frac{\pi}{4}[/tex] - cos [tex]\frac{\pi}{6}[/tex] = [tex]\frac{2\sqrt{2}-\sqrt{3}} {2}[/tex]
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Answer: secθ
Step-by-step explanation:
cosθ csc²θ tan²θ
= cosθ * [tex]\frac{1}{sin^{2}\theta}[/tex] * [tex]\frac{sin^{2}\theta} {cos^{2}\theta}[/tex]
= [tex]\frac{1}{cos\theta}[/tex]
= secθ
