Question

In ΔABC, if C = 90° then prove that $(\frac{a^{2} + b^{2}}{a^{2} - b^{2}}) sin(A-B) = 1$

Answer 1

Answer:

Step-by-step explanation:

Formula used:

In triangle ABC,

(a/SinA)=(b/sinB)=(c/sinC)=2R

${sin}^2A-{sin}^2B=sin(A+B).sin((A-B)$

In triangle ABC, A+B+C=180

But C=90, A+B=90

$(\frac{{a}^2+{b}^2}{{a}^2-{b}^2}).sin(A-B)$

$\frac{{(2R.sinA)}^2+{(2R.sinB)}^2}{{(2R.sinA)}^2-{(2R.sinA)}^2}.sin(A-B)$

$\frac{{2R}^2[{sin}^2A+{sin}^2B]}{{2R}^2[{sin}^2A-{sin}^2B]}.sin(A-B)$$\:\:=\frac{[{sin}^2A+{sin}^2B]}{[{sin}^2A-{sin}^2B]}.sin(A-B)$

$\frac{[{sin}^2A+{sin}^2(90-A)]}{sin(A+B).sin(A-B)}.sin(A-B)$

$\frac{[{sin}^2A+{cos}^2A]}{sin(90)}$

$\frac{1}{1}\:=\:1$