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Question Number 135065 by liberty last updated on 10/Mar/21

$$\mathrm{cosec}{}^{2}68°+\sec {}^{2}56°-\cot {}^{2}34°-\tan {}^{2}22°=?$$

Commented by som(math1967) last updated on 10/Mar/21

$$2$$

Commented by som(math1967) last updated on 10/Mar/21

$${cosec}^{2}68-{cot}^2(90-22)$$$$+{sec}^{2}56-{tan}^2(90-34)$$$${cosec}^{2}68-{cot}^{2}68+{sec}^{2}56-{tan}^{2}56$$$$=1+1=2$$

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