1+\(cot^2\)x=\(\dfrac{1}{sin^2x}\)\(\Leftrightarrow\)1+\(\dfrac{3}{4}^2\)=\(\dfrac{1}{sin^2x}\)➩\(sin^2x\)=\(\dfrac{16}{25}\)\(\Rightarrow\)sinx=\(\dfrac{4}{5}\)
\(\sin^2x+\cos^2x=1\)\(\Rightarrow\)cosx=\(\sqrt{1-sin^2x}\)=\(\dfrac{3}{5}\)
tanx=\(\dfrac{\sin x}{\cos x}\)=\(\dfrac{4}{3}\)