Question

Prove that tan50+cot50=2sec10

Answer 1

Hope this help..........

Answer 2

$LHS = tan 50 \degree + cot 50 \degree \\= tan 50 \degree + \frac{1}{tan 50 \degree }\\= \frac{ tan^{2} 50 \degree + 1 }{tan 50 \degree } \\= \frac{sec^{2} 50 \degree }{tan 50 \degree}$

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$\blue { By \: Trigonometric \:Identity )}$

$\pink { 1 + tan^{2} \theta = sec^{2} \theta }$

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$= \frac{\frac{1}{cos^{2} 50 }}{ \frac{sin 50}{cos 50 }} \\= \frac{1}{ sin 50 cos 50 } \\= \frac{2}{2 sin 50 cos 50 } \\= \frac{2}{ sin ( 2\times 50 )}$

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We know that,

$\orange { 2sinA cos A =sin 2A }$

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$= \frac{2}{sin 100} \\= 2 Cosec 100 \\= 2 Cosec (90 + 10 ) \\= 2sec10 \\= RHS$

$Hence \:proved$

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