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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:
$\cos 30^\circ . \cos 60^\circ - \sin 30^\circ . \sin 60^\circ = 0$
✓

Answer

$\text{LHS}=\cos30^\circ .\cos 60^\circ - \sin 30^\circ . \sin 60^\circ $
$=\frac{\sqrt{3}}{2} \frac{1}{2}-\frac{1}{2} \frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$=0$
$=\text{RHS}$

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