In $Δ A B C,$ if ${sin}^{2} A + {sin}^{2} B = {sin}^{2} C,$ then show that the triangle is a right angled triangle.

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Solution

${sin}^{2} A + {sin}^{2} B = {sin}^{2} C$
${sin}^{2} A = {sin}^{2} C - {sin}^{2} B$
${sin}^{2} A = sin (C - B) sin (C + B)$
as $sin A = sin (π - (B + C)) = sin (B + C)$
${sin}^{2} (B + C) = sin (C - B) sin (C + B)$
$sin (C + B) = sin (C - B)$
Thus $C + B = π - (C - B); C + B \neq C - B$ because then $B$ will become $0$
$C = \frac{π}{2}$
Therefore triangle is right angled
$sin (α - β) (sin α + β)$
$(sin α cos β)^{2}$ - $(sin β - cos α)^{2}$
${sin}^{2} α - {sin}^{2} β$

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