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Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSTrigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]2 Marks
Question
Prove that:$3\operatorname{cosec}^2 60^\circ - 2 \cot^2 30^\circ + \sec^2 45^\circ = 0$
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Answer

$LHS =3 \operatorname{cosec}^260^\circ – 2 \cot^230^\circ + \sec^245^\circ$
$=3\left(\frac{2}{\sqrt{3}}\right)^2-2(\sqrt{3})^2+(\sqrt{2})^2$
$=3 \times \frac{4}{3}-2 \times 3+2$
$=4-6+2$
$=0$
$=\text { RHS }$

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