Question

assume that 0<x<pi/2 and 0<y<pi/2. find the exact value of tan(x-y) if sin x=8/17 and cos y= 3/5

Answer 1

Recall that

$\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$

$\cos^2\theta+\sin^2\theta=1$

From the second identity, we find

$\cos x=\pm\sqrt{1-\sin^2x}$

$\sin y=\pm\sqrt{1-\cos^2y}$

We're told that $0<x<\dfrac\pi2$ and $0<y<\dfrac\pi2$, which means we should take the positive roots above. Then

$\cos x=\sqrt{1-\left(\dfrac8{17}\right)^2}=\dfrac{15}{17}$

$\sin y=\sqrt{1-\left(\dfrac35\right)^2}=\dfrac45$

Then we compute the tangents:

$\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac8{17}}{\frac{15}{17}}=\dfrac8{15}$

$\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac45}{\frac35}=\dfrac43$

so we end up with

$\tan(x-y)=\dfrac{\frac8{15}-\frac43}{1+\frac8{15}\cdot\frac43}=-\dfrac{36}{77}$

Answer 2

The exact value of $tan(x-y)=-\frac{36}{77}$

We have $sinx=\frac{8}{17} ,cosy=\frac{3}{5}$ and $0<x<\frac{\pi}{2} , 0<y<\frac{\pi}{2}$

That is the trigonometric functions lies in the first quadrant .

As,

$tanx=\frac{sinx}{cosx} \\=\frac{sinx}{\sqrt{1-sin^{2} x} }\\\frac{8}{\sqrt{17^{2}-8^{2} } } \\=\frac{8}{\sqrt{225} } \\=\frac{8}{15} \\tany=\frac{siny}{cosy} \\\\=\frac{siny}{\sqrt{1-sin^{2}y} } =\frac{\sqrt{5^{2}-3^{2} } }{3} \\=\frac{4}{3}$

Now, we compute :

$tan(x-y)=\frac{tanx-tany}{1+tanx.tany} \\=\frac{\frac{8}{15}-\frac{4}{3} }{1+{\frac{8}{15}.\frac{4}{3}}} \\tan(x-y)=-\frac{36}{77}$

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