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$C o t (1358^{0}) + \tan (3608^{0}) =$

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Correct option is B

cot(1358)=cot(15×90+8)=-tan8 tan3608=tan(40×90+8)=tan8 adding both

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Cot(13580)+tan(36080)=

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$C o t (1358^{0}) + \tan (3608^{0}) =$