Prove that $\tan 225 ° c o t 405 ° + \tan 765 ° c o t 675 ° = 0 .$

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Solution

Determine the proving of the expression $\tan 225 ° c o t 405 ° + \tan 765 ° c o t 675 ° = 0 .$
Solve L.H.S part:
$\tan 225 ° c o t 405 ° + \tan 765 ° c o t 675 ° = \tan (90 ° \times 2 + 45 °) c o t (90 ° \times 4 + 45 °) + \tan (90 ° \times 8 + 45 °) c o t (90 ° \times 7 + 45 °) \Rightarrow = \tan 45 ° c o t 45 ° + \tan 45 ° [- \tan 45 °] \Rightarrow = \tan 45 ° c o t 45 ° - \tan 45 ° \tan 45 ° ∵ c o t 45 ° = t a n 45 ° = 1 \Rightarrow = 1 \times 1 - 1 \times 1 \Rightarrow = 1 - 1 \Rightarrow = 0$
Hence, the L.H.S=R.H.S.

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tan (- 225^{o}) cot (- 405^{o}) - tan (- 765^{0}) cot (675^{o}) = 0

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tan 225^{o} cot 405^{o} + tan 765^{o} = 2

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