a) $cos\alpha=0,8$

Ta có: $sin^2\alpha+cos^2\alpha=1$

        → $sin^2\alpha+(0,8)^2=1$

        → $sin^2\alpha=\dfrac{9}{25}$

        → \(\left[ \begin{array}{l}sin\alpha=\dfrac{3}{5}\\sin\alpha=-\dfrac{3}{5}\end{array} \right.\) 

+) Với giá trị $sin\alpha=\dfrac{3}{5}$

→ $tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{5}÷0,8=\dfrac{3}{4}$

→ $cot\alpha=\dfrac{1}{tan\alpha}=1÷\dfrac{3}{4}=\dfrac{4}{3}$

+) Với giá trị $sin\alpha=-\dfrac{3}{5}$

→ $tan\alpha=\dfrac{sin\alpha}{cos\alpha}=-\dfrac{3}{5}÷0,8=-\dfrac{3}{4}$

→ $cot\alpha=\dfrac{1}{tan\alpha}=1÷(-\dfrac{3}{4})=\dfrac{4}{3}$

b) $tan\alpha=\dfrac{3}{4}$

→ $cot\alpha=\dfrac{4}{3}$

Ta có: $1+tan^2\alpha=\dfrac{1}{cos^2\alpha}$

      ⇒ $1+(\dfrac{3}{4})^2=\dfrac{1}{cos^2\alpha}$

      ⇔ $cos^2\alpha=\dfrac{16}{25}$

      ⇔ \(\left[ \begin{array}{l}cos\alpha=\dfrac{4}{5}\\cos\alpha=-\dfrac{4}{5}\end{array} \right.\) 

+) Với giá trị $cos\alpha=\dfrac{4}{5}$

→ $sin\alpha=tan\alpha×cos\alpha=\dfrac{3}{4}×\dfrac{4}{5}=\dfrac{3}{5}$

+) Với giá trị $cos\alpha=-\dfrac{4}{5}$

→ $sin\alpha=tan\alpha×cos\alpha=\dfrac{3}{4}×\dfrac{4}{5}=-\dfrac{3}{5}$

c) $cot\alpha=\dfrac{\sqrt{7}}{3}$

→ $tan\alpha=\dfrac{3\sqrt{7}}{7}$

Ta có: $1+cot^2\alpha=\dfrac{1}{sin^2\alpha}$

        ⇒ $1+(\dfrac{\sqrt{7}}{3})^2=\dfrac{1}{sin^2\alpha}$

        ⇔ $sin^2\alpha=\dfrac{9}{16}$

        ⇔ \(\left[ \begin{array}{l}sin\alpha=\dfrac{3}{4}\\sin\alpha=-\dfrac{3}{4}\end{array} \right.\) 

+) Với giá trị $sin\alpha=\dfrac{3}{4}$

→ $cos\alpha=cot\alpha×sin\alpha=\dfrac{\sqrt{7}}{3}×\dfrac{3}{4}=\dfrac{\sqrt{7}}{4}$

+) Với giá trị $sin\alpha=-\dfrac{3}{4}$

→ $cos\alpha=cot\alpha×sin\alpha=\dfrac{\sqrt{7}}{3}×(-\dfrac{3}{4})=-\dfrac{\sqrt{7}}{4}$

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