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Prove that :

tan \(\cfrac{5\pi}4\) cot \(\cfrac{9\pi}4\) + tan \(\cfrac{17\pi}4\) cot \(\cfrac{15\pi}4\) = 0 

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LHS = tan \(\cfrac{5\pi}4\) cot \(\cfrac{9\pi}4\) + tan \(\cfrac{17\pi}4\) cot \(\cfrac{15\pi}4\) 

= tan 225° cot 405° + tan 765° cot 675°

= tan (90° × 2 + 45°) cot (90° × 4 + 45°) + tan (90° × 8 + 45°) cot (90° × 7 + 45°)

We know that when n is odd, cot → tan.

= tan 45° cot 45° + tan 45° [-tan 45°]

= tan 45° cot 45° - tan 45° tan 45°

= 1 × 1 – 1 × 1

= 1 – 1

= 0

= RHS

Hence proved.

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