Question Number 32891 by mondodotto@gmail.com last updated on 05/Apr/18 | ||
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Commented by Tinkutara last updated on 05/Apr/18 | ||
It is wrong. | ||
Commented by Joel578 last updated on 05/Apr/18 | ||
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$$\mathrm{sec}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}} \\ $$ | ||
Commented by mrW2 last updated on 06/Apr/18 | ||
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$$\mathrm{sec}^{−\mathrm{1}} {x}\neq\left(\mathrm{sec}\:{x}\right)^{−\mathrm{1}} \\ $$$$\mathrm{sec}^{−\mathrm{1}} {x}={arcsec}\:{x} \\ $$$$\left({sec}\:{x}\right)^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{sec}\:{x}}=\mathrm{cos}\:{x} \\ $$ | ||