Question

If $\tan \left(\mathrm{A}\right)=\cot \left(\mathrm{B}\right)$ then prove that $\mathrm{A}+\mathrm{B}=90°$.

Answer 1

Using basic trigonometric Identities

Using the fact that $\tan \left(\mathrm{\theta }\right)$ and $\cot \left(\mathrm{\theta }\right)$ are complementary Trigonometric Ratios.

$$\begin{gathered} \cot \left(\mathrm{B}\right)=\tan \left(90°-\mathrm{B}\right) \\ \Rightarrow \tan \left(90°-\mathrm{B}\right)=\tan \left(\mathrm{A}\right) \\ \Rightarrow 90°-\mathrm{B}=\mathrm{A} \\ \Rightarrow \mathrm{A}+\mathrm{B}=90° \end{gathered}$$

Hence Proved.