Question

For an acute angle theta,bsin theta=cos theta. Therefore,. a sin theta - b cos theta

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a sin theta +b cos theta

Answer 1

Answer:

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

Step-by-step explanation:

Given:

$b\:sin\theta=a\:cos\theta$

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

$sin\theta:cos\theta$= a : b

$sin\theta = ak \\\\cos\theta= bk$

Now,

$\frac{a\:sin\theta-b\:cos\theta}{a\:sin\theta+b\:cos\theta}\\\\=\frac{a(ak)-b(bk)}{a(ak)+b(bk)}\\\\=\frac{a^2k-b^2k}{a^2k+b^2k}\\\\=\frac{k(a^2-b^2)}{k(a^2+b^2)}\\\\=\frac{a^2-b^2}{a^2+b^2}$