If $\tan A + \cot A = 4$,then $\tan^4 A + \cot^4 A$ is equal to
- A$110$
- B$191$
- C$80$
- D$194$
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Medium
View Solution$\frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2} = \dots$
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