Question

Prove that following
$\sin {40}^{o} - \cos {70}^{o} = \sqrt{3} \cos {80}^{o}$
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Answer 1

Prove that following

$\sin {40}^{o} - \cos {70}^{o} = \sqrt{3} \cos {80}^{o}$

$\rule{150}2$

○ sOLUTIOn ○

L.H.S=

$\leadsto sin40°-cos70° \\ \leadsto sin40°-cos(90°-70°)$

Using identity

$\red{\fbox{sinA-sinB=\frac{2cos(A+B)}{2}×\frac{2sin(A-B)}{2} }}$

Hence,

$\leadsto sin40°-sin20°=2cos30°×sin10°\\ \leadsto 2×\frac{\sqrt3}{2}×cos80°\\ \leadsto \sqrt3\: cos\:80°\:R.H.S$

Proved