✨ Best Answer ✨
{cosθ/(1+sinθ)}+tanθ={cosθ/(1+sinθ)}+{sinθ/cosθ}
={cos²θ+sinθ+sin²θ}/{cosθ(1+sinθ)}
={1+sinθ}/{cosθ(1+sinθ)}
=1/cosθ
したがって、
{cosθ/(1+sinθ)}+tanθ=1/cosθ
例題4のように問12を解くと式はどうなりますか
✨ Best Answer ✨
{cosθ/(1+sinθ)}+tanθ={cosθ/(1+sinθ)}+{sinθ/cosθ}
={cos²θ+sinθ+sin²θ}/{cosθ(1+sinθ)}
={1+sinθ}/{cosθ(1+sinθ)}
=1/cosθ
したがって、
{cosθ/(1+sinθ)}+tanθ=1/cosθ
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