If tantheta = a/b, then how do you show that (asintheta - bcostheta)/(asintheta + bcostheta) = (a^2 - b^2)/(a^2 + b^2)?

Feb 11, 2017

We know that $\tan \theta = \sin \frac{\theta}{\cos} \theta$, so we know that $\sin \theta = a$ and $\cos \theta = b$

$\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} = \frac{a \left(a\right) - b \left(b\right)}{a \left(a\right) + b \left(b\right)}$

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

Hopefully this helps!

Feb 11, 2017

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

Explanation:

We know $\tan \theta = \frac{a}{b} = \frac{h e i g h t}{b a s e}$

In a right angle triangle, we know, ${\left(h y p o t e \nu s e\right)}^{2} = {\left(h e i g h t\right)}^{2} + {\left(b a s e\right)}^{2}$

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

Now sin$\theta = \frac{h e i g h t}{h y p o t e \nu s e} = \frac{a}{\sqrt{{a}^{2} + {b}^{2}}}$ and
cos $\theta = \frac{b a s e}{h y p o t e \nu s e} = \frac{b}{\sqrt{{a}^{2} + {b}^{2}}}$

So, $\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta}$

= [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)}]

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

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

= R.H.S

Feb 12, 2017

$\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta}$

$= \frac{a \sin \frac{\theta}{\cos} \theta - b \cos \frac{\theta}{\cos} \theta}{a \sin \frac{\theta}{\cos} \theta + b \cos \frac{\theta}{\cos} \theta}$

$= \frac{a \tan \theta - b}{a \tan \theta + b}$

$= \frac{a \times \frac{a}{b} - b}{a \times \frac{a}{b} + b}$

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

Proved