Question

solve tan(pi/4 + A) + cot(pi/4 + A) = 4, then A=

Answer 1

$$\begin{gathered} \tan \left(\frac{\mathrm{\pi }}{4}+A\right)+\cot \left(\frac{\mathrm{\pi }}{4}+A\right)=4 \\ \Rightarrow \frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+A\right)}{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}+A\right)}+\frac{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}+A\right)}{\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+A\right)}=4 \\ \Rightarrow \frac{{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}+A\right)+{\cos }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}+A\right)}{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}+A\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+A\right)}=4 \\ \Rightarrow \frac{1}{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}+A}\right)\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}+A}\right)}=4 \\ \Rightarrow 4\cos \left(\frac{\mathrm{\pi }}{4},+,A\right)\sin \left(\frac{\mathrm{\pi }}{4},+,A\right)=1 \\ \Rightarrow 2\times 2\sin \left(\frac{\mathrm{\pi }}{4},+,A\right)\cos \left(\frac{\mathrm{\pi }}{4},+,A\right)=1 \\ \Rightarrow 2\sin \left[2\left(\frac{\mathrm{\pi }}{4},+,A\right)\right]=1 \\ \Rightarrow \sin \left[\frac{\mathrm{\pi }}{2}+2A\right]=\frac{1}{2} \\ \Rightarrow \cos \left(2A\right)=\frac{1}{2} \\ \Rightarrow \cos \left(2A\right)=\cos \left(\frac{\mathrm{\pi }}{3}\right) \\ \therefore 2A=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3} \\ \mathrm{Hence},A=n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6} \end{gathered}$$