Question

How cot[180⁰-@]=-cot@

Answer 1

$$\begin{gathered} \cot \left(180°-\mathrm{\theta }\right) \\ =\frac{\cos \left(180°-\mathrm{\theta }\right)}{\sin \left(180°-\mathrm{\theta }\right)} \\ \mathrm{Now},\cos \left(180°-\mathrm{\theta }\right)=\cos 180°.\cos \mathrm{\theta }+\sin 180°.\sin \mathrm{\theta }=-1.\cos \mathrm{\theta }+0.\sin \mathrm{\theta }=-\cos \mathrm{\theta } \\ \mathrm{Now},\sin \left(180°-\mathrm{\theta }\right)=\sin 180°.\cos \mathrm{\theta }-\cos 180°.\sin \mathrm{\theta }=0.\cos \mathrm{\theta }-\left(-1\right).\sin \mathrm{\theta }=\sin \mathrm{\theta } \\ \mathrm{Now},\cot \left(180°-\mathrm{\theta }\right)=\frac{\cos \left(180°-\mathrm{\theta }\right)}{\sin \left(180°-\mathrm{\theta }\right)}=\frac{-\cos \mathrm{\theta }}{\sin \mathrm{\theta }}=-\cot \mathrm{\theta } \end{gathered}$$