Question

How to solve this :') please help

Answer 1

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

$\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}$

Use $\cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1$

so

$\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}$