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Trigonometric Ratios of Complementary Angles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Ratios of Complementary Angles1 Mark
Question
Without using trignometric tables, evaluate:
$\frac{\tan27^\circ}{\cot63^\circ}$
✓

Answer

$\frac{\tan27^\circ}{\cot63^\circ}$
$=\frac{\tan(90^\circ-63^\circ)}{\cot63^\circ}$
$=\frac{\cot63^\circ}{\cot63^\circ}$ $[\therefore\ \tan(90-\theta)=\cot\theta]$
$=1$

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