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Question

${sin}^{2} 62^{\circ} - {cos}^{2} 28^{\circ}$

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Solution

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${sin}^{2} 62^{\circ} - {cos}^{2} 28^{\circ}$
= ${sin}^{2} (90 - 28)^{\circ} - {cos}^{2} 28^{\circ}$
= ${cos}^{2} 28^{\circ} - {cos}^{2} 28^{\circ}$
= $0$

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Q1

{sin}^{2} 62^{\circ} - {cos}^{2} 28^{\circ}

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Q2

(cos²28° − sin²62°) = ?
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\frac{(\sin^{2} 73 ° + \sin^{2} 17 °)}{(\cos^{2} 28 ° + \cos^{2} 62 °)} = 1 .

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