Triangles And Properties Of Triangle Question 27
In a triangle ABC, $ \angle DC=90{}^\circ , $ then $ \frac{a^{2}-b^{2}}{a^{2}+b^{2}} $ is equal to:
Options:
A) $ \sin (A+B) $
B) $ \sin (A-B) $
C) $ \cos (A+B) $
D) $ \sin ( \frac{A-B}{2} ) $
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Answer:
Correct Answer: A
Solution:
[b] $ A+B=180{}^\circ -C=90{}^\circ $ $ a=2R\sin A,b=2R\sin B,c=2R\sin C $
$ \therefore ,\frac{a^{2}-b^{2}}{a^{2}+b^{2}}=\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{{{\sin }^{2}}A+{{\sin }^{2}}B} $ $ =\frac{\sin (A+B)\sin(A-B)}{{{\sin }^{2}}A+{{\sin }^{2}}(90{}^\circ -A)} $ $ [\because ,A+B=90{}^\circ ] $ $ =\frac{\sin 90{}^\circ \sin (A-B)}{{{\sin }^{2}}A+{{\cos }^{2}}A}=\sin (A-B) $
BETA
