Question
Download Solution PDFFind the value of \(\frac{{{{\tan }^2}60^\circ + {{\cot }^2}30^\circ }}{{{{\tan }^2}20^\circ - {{\sec }^2}20^\circ }} - \frac{{\sin 90^\circ + \cos 0^\circ }}{{\cos e{c^2}20^\circ - {{\cot }^2}20^\circ }}\).
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFFormula used
tan 60º = √3
cot 30º = √3
sec2a - tan2a = 1
cosec2a - cot2a = 1
Calculation
\(\frac{{{{\tan }^2}60^\circ + {{\cot }^2}30^\circ }}{{{{\tan }^2}20^\circ - {{\sec }^2}20^\circ }} - \frac{{\sin 90^\circ + \cos 0^\circ }}{{\cos e{c^2}20^\circ - {{\cot }^2}20^\circ }}\)
⇒ (√32 + √32)/(-1) - (1 + 1)/1
⇒ - 6 - 2 = - 8
The answer is -8.
Last updated on Jun 2, 2025
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