Show that $s i n 10 ° - c o s 10 ° s i n 80 ° c o s 80 ° = 1$

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Solution

Consider, LHS: $sin 10 ° - cos 10 ° sin 80 ° cos 80 °$
$= sin 10 ° cos 80 ° + cos 10 ° sin 80 °$
$= sin 10 ° cos (90 ° - 10 °) + cos 10 ° sin (90 ° - 10 °)$
[Since, $c o s θ = sin (90 - θ)$ ]
$= sin 10 ° sin 10 ° + cos 10 ° cos 10 °$
$= {sin}^{2} 10 ° + {cos}^{2} 10 °$ [Since, ${sin}^{2} θ + {cos}^{2} θ = 1$ ]
$= 1$
$=$ RHS

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