Problem 901

1. In the numerator of the first fraction, we use the formula for the difference of squares and write the product $\left(1-\cos ^{2} x\right)\left(1+\cos ^{2} x\right)$. We consider the following relationships: $\tan x=\frac{\sin x}{\cos x}$ and $1-\cos ^{2} x=\sin ^{2} x$. We simplify the first fraction by $\mathrm{s} \sin ^{2} x$ and raise the second fraction to the exponent -2. We add the fractions and get $\frac{1-\cos ^{2} x}{\sin ^{2} x}$. In the numerator, we again consider the relationship $1-\cos ^{2} x=\sin ^{2} x$, so the result is 1.

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