Question

If $tanA+cotA=2$, then find the value of ${\tan }^{2}A+{\cot }^{2}A$.

Answer 1

It is given that $\tan A+\cot A=2$, squaring both sides, we get:

$(\tan A+\cot A{)}^2=2^2\Rightarrow \tan {}^{2}A+\cot {}^{2}A+(2\times \tan A\times \cot A)=4(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow \tan {}^{2}A+\cot {}^{2}A+(2\times \tan A\times \frac{1}{\tan A})=4(\because \cot x=\frac{1}{\tan x})\Rightarrow \tan {}^{2}A+\cot {}^{2}A+2=4\Rightarrow \tan {}^{2}A+\cot {}^{2}A=4-2\Rightarrow \tan {}^{2}A+\cot {}^{2}A=2$

Hence, $\tan {}^{2}A+\cot {}^{2}A=2$.