Question
Evaluate the following$: \sec16^\circ \tan28^\circ - \cot62^\circ \operatorname{cosec}74^\circ $
✓
Answer
$\sec16° \tan28° - \cot62° \operatorname{cosec}74°$
$= \sec(90° - 74°)\tan(90° - 62°) - \cot62° \operatorname{cosec}74°$
$= \operatorname{cosec}74° \cot62° - \cot62° \operatorname{cosec}74°$
$= 0$
$= \sec(90° - 74°)\tan(90° - 62°) - \cot62° \operatorname{cosec}74°$
$= \operatorname{cosec}74° \cot62° - \cot62° \operatorname{cosec}74°$
$= 0$
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