$sec\left(40\right)-\tan\left(40\right)$
$\frac{1}{2}+\frac{1}{2^3}4$
$12+4+2-13-24-2+3-2$
$\lim_{x\to\infty}15x\cdot tan\frac{12}{x}$
$4x-2x+1x-9x$
$-5\sqrt{2x}+8\sqrt{2x}$
$\sec\theta\csc\theta-\tan\theta\equiv\cot\theta$
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