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Trigonometric Ratios of Complementary Angles — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsTrigonometric Ratios of Complementary Angles1 Mark
Question
Without using trignometric tables, prove that:
$\text{cosec}80^\circ-\sec10^\circ=0$
✓

Answer

$\text{L.H.S.}=\text{cosec}80^\circ-\sec10^\circ$
$=\text{cosec}\big(90^\circ-10^\circ\big)-\sec10^\circ$
$=\sec10^\circ-\sec10^\circ$
$=0$
$=\text{R.H.S}$

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