Question

please solve the problem wrong answer will be reported salute ​

Answer 1

$LHS = cot 17 \degree \Big( cot 73 \degree cos^{2} 22\degree + \frac{1}{tan 73\degree sec^{2} 68\degree }\Big)$

$= cot (90 - 73) \degree \Big( cot 73 \degree cos^{2} 22\degree + \frac{1}{tan 73\degree sec^{2} (90 - 22)\degree} \Big)$

$= tan 73 \degree \Big( cot 73 \degree cos^{2} 22\degree + \frac{1}{tan 73\degree cosec^{2} 22\degree }\Big)$

$= tan 73 \degree cot 73 \degree cos^{2} 22\degree + \frac{\cancel {tan 73 \degree}}{\cancel {tan 73\degree }cosec^{2} 22\degree} \Big)$

$= cos^{2} 22\degree + sin^{2} 22\degree$

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$1. tan A \times cot A = 1 \\</p><p>2. \frac{1}{Cosec A } = sin A$

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$= 1$

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/* By Trigonometric Identity */

• Sin² A + cos² A = 1

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$= RHS$

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