問題文全文(内容文)：
$\tan(\cos^{-1}x)=\sin(\tan^{-1}\displaystyle \frac{4}{7})$を満たす$x$を求めよ。
$0 \leqq \cos^{-1}x \leqq \pi$
$-\displaystyle \frac{1}{2} \lt \tan^{-1}x \lt \displaystyle \frac{\pi}{2}$
$\tan(\cos^{-1}x)=\sin(\tan^{-1}\displaystyle \frac{4}{7})$を満たす$x$を求めよ。
$0 \leqq \cos^{-1}x \leqq \pi$
$-\displaystyle \frac{1}{2} \lt \tan^{-1}x \lt \displaystyle \frac{\pi}{2}$
単元：
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指導講師：
ますただ
問題文全文(内容文)：
$\tan(\cos^{-1}x)=\sin(\tan^{-1}\displaystyle \frac{4}{7})$を満たす$x$を求めよ。
$0 \leqq \cos^{-1}x \leqq \pi$
$-\displaystyle \frac{1}{2} \lt \tan^{-1}x \lt \displaystyle \frac{\pi}{2}$
$\tan(\cos^{-1}x)=\sin(\tan^{-1}\displaystyle \frac{4}{7})$を満たす$x$を求めよ。
$0 \leqq \cos^{-1}x \leqq \pi$
$-\displaystyle \frac{1}{2} \lt \tan^{-1}x \lt \displaystyle \frac{\pi}{2}$
投稿日：2021.11.27