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$\frac{3 + \cot 76 ° \cot 16 °}{\cot 76 ° + \cot 16 °}$ is equal to

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tan16°

tan76°

tan44°

tan46°

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detailed solution

Correct option is D

3+cot76°cos16°cot76°+cot16°=3sin76°sin16°+cos76°cos16°sin76°+16°=2sin76°sin16°+cos76°−16°sin92°=cos60°−cos92°+cos60°sin92°=1−cos92°sin92°=2sin246°2sin46°cos46°=tan46°=tan90°−44°=cot44°

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3+cot76° cot16°cot76°+cot16°is equal to

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$\frac{3 + \cot 76 ° \cot 16 °}{\cot 76 ° + \cot 16 °}$ is equal to