정답

(1) \(\displaystyle \cos h ^{2} x-\sin  h ^{2} x= \left (  \frac {e ^{x} +e ^{-x}} {2} \right) ^{2} - \left (  \frac {e ^{x} -e ^{-x}} {2} \right) ^{2} =1 \)

(2) \(\displaystyle  ( \sin  hx) ' = \left (  \frac {e ^{x} -e ^{-x}} {2} \right) ' = \left (  \frac {e ^{x} +e ^{-x}} {2} \right) =\cos hx \)

\(\displaystyle  ( \cos hx) ' = \left (  \frac {e ^{x} +e ^{-x}} {2} \right) ' = \left (  \frac {e ^{x} -e ^{-x}} {2} \right) =\sin  hx \)

이므로 

\(\displaystyle \tan h ' x= \left (  \frac {\sin  hx} {\cos hx} \right) ^{' } = \frac {\cos h ^{2} x-\sin  h ^{2} x} {\cos h ^{2} x} = \frac {1} {\cos h ^{2} x} =\sec h ^{2} x \)

\(\displaystyle \therefore  ~\tan h ' x=\sec h ^{2} x \)

\(\displaystyle \int   _{} ^{} {\tan h ^{2} x} dx= \int   _{} ^{} {1-\sec h ^{2} x} dx=x-\tan hx+C \)