If $tan θ = \frac{a}{b}$ $tantheta = a/b$ , then how do you show that $\frac{a sin θ - b cos θ}{a sin θ + b cos θ} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ $(asintheta - bcostheta)/(asintheta + bcostheta) = (a^2 - b^2)/(a^2 + b^2)$ ?

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If $tan θ = \frac{a}{b}$ $tantheta = a/b$ , then how do you show that $\frac{a sin θ - b cos θ}{a sin θ + b cos θ} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ $(asintheta - bcostheta)/(asintheta + bcostheta) = (a^2 - b^2)/(a^2 + b^2)$ ?

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Answer 1 · 2017-02-11T15:16:26

We know that $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sintheta/costheta$ , so we know that $sin θ = a$ $sintheta = a$ and $cos θ = b$ $costheta = b$

$\frac{a sin θ - b cos θ}{a sin θ + b cos θ} = \frac{a (a) - b (b)}{a (a) + b (b)}$ $(asintheta - bcostheta)/(asintheta + bcostheta) = (a(a) - b(b))/(a(a) + b(b))$

This can be written as $\frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ $(a^2 - b^2)/(a^2 + b^2)$ .

Hopefully this helps!

Answer 2 · 2017-02-11T15:47:53

proved R.H.S = L.H.S

Explanation:

We know $tan θ = \frac{a}{b} = \frac{h e i g h t}{b a s e}$ $tan theta = a/b =[height]/[base]$

In a right angle triangle, we know, ${(h y p o t e ν s e)}^{2} = {(h e i g h t)}^{2} + {(b a s e)}^{2}$ $(hypotenuse)^2 = (height)^2 +(base)^2$

So, hypotenuse = $\sqrt{a^{2} + b^{2}}$ $sqrt (a^2 + b^2)$

Now sin $θ = \frac{h e i g h t}{h y p o t e ν s e} = \frac{a}{\sqrt{a^{2} + b^{2}}}$ $theta = (height)/(hypotenuse) = a/[sqrt(a^2 +b^2)]$ and
cos $θ = \frac{b a s e}{h y p o t e ν s e} = \frac{b}{\sqrt{a^{2} + b^{2}}}$ $theta = (base)/(hypotenuse) =b/[sqrt(a^2 + b^2)]$

So, $\frac{a sin θ - b cos θ}{a sin θ + b cos θ}$ $[a sin theta - b cos theta]/[a sin theta + b cos theta]$

= #[a.{a/sqrt(a^2+b^2)} - [b{b/sqrt(a^2+b^2)}]]/[[a{a/sqrt(a^2+b^2)}] +[b{b/sqrt(a^2+b^2)}]#

= $\frac{\frac{a^{2}}{\sqrt{a^{2} + b^{2}}} - \frac{b^{2}}{\sqrt{a^{2} + b^{2}}}}{\frac{a^{2}}{\sqrt{a^{2} + b^{2}}} - \frac{b^{2}}{\sqrt{a^{2} + b^{2}}}}$ $[a^2/sqrt(a^2+b^2)- b^2/sqrt(a^2+b^2)]/[a^2/sqrt(a^2+b^2)-b^2/sqrt(a^2+b^2)]$

= $\frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ $(a^2 - b^2)/(a^2+b^2)$

= R.H.S

Answer 3 · 2017-02-12T18:04:57

$\frac{a sin θ - b cos θ}{a sin θ + b cos θ}$ $(asintheta - bcostheta)/(asintheta + bcostheta)$

$= \frac{a \frac{sin θ}{cos θ} - b \frac{cos θ}{cos θ}}{a \frac{sin θ}{cos θ} + b \frac{cos θ}{cos θ}}$ $=(asintheta/costheta - bcostheta/costheta)/(asintheta/costheta+ bcostheta/costheta)$

$= \frac{a tan θ - b}{a tan θ + b}$ $=(atantheta - b)/(atantheta+ b)$

$= \frac{a \times \frac{a}{b} - b}{a \times \frac{a}{b} + b}$ $=(axxa/b - b)/(axxa/b+ b)$

$= \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ $=(a^2 - b^2)/(a^2+ b^2)$

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If $tan θ = \frac{a}{b}$ $tantheta = a/b$ , then how do you show that $\frac{a sin θ - b cos θ}{a sin θ + b cos θ} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$ $(asintheta - bcostheta)/(asintheta + bcostheta) = (a^2 - b^2)/(a^2 + b^2)$ ?

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Explanation:

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