## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

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Recall the co-function identity: $\sec (\theta)=\csc(90^{\circ}-\theta)$ We know that $1+\tan^2 \theta=\sec^2 \theta$ Therefore, $1-\tan^2 \ (5^{\circ})-\csc^2 (70^{\circ}) =\sec^2 \ (5^{\circ})-\csc^2 (85^{\circ}) \\=\csc^2 \ (90^{\circ}-85^{\circ})- \csc^2 (85^{\circ}) \\=\csc^2 \ (85^{\circ})- \csc^2 (85^{\circ}) \\=0$