Question

Answer the question in the picture

Answer 1

Recall the angle sum identities:

`\sin(x+y)=\sin x\cos y+\cos x\sin y`

`\cos(x+y)=\cos x\cos y-\sin x\sin y`

Now,

`\tan(x+y)=(\sin(x+y))/(\cos(x+y))=(\sin x\cos y+\cos x\sin y)/(\cos x\cos y-\sin x\sin y)`

Divide through numerator and denominator by
`\cos x\cos y` to get

`\tan(x+y)=(\tan x+\tan y)/(1-\tan x\tan y)`

Next, we use the fact that
`x,y` lie in the first quadrant to determine that

`\sin x=\frac12\implies\cos x=√(1-\sin^2x)=\frac{\sqrt3}2`

`\cos y=\frac{\sqrt2}2\implies\sin x=√(1-\cos^2x)=\frac1{\sqrt2}`

So we then have

`\tan x=(\sin x)/(\cos x)=\frac{\frac12}{\frac{\sqrt3}2}=\frac1{\sqrt3}`

`\tan y=(\sin y)/(\cos y)=\frac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1`

Finally,

`\tan(x+y)=\frac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=(1+\sqrt3)/(\sqrt3-1)=2+\sqrt3\approx3.73`