Question Number 59683 by ranusahu last updated on 13/May/19
$${prove}\:\left(\mathrm{1}+{tanx}\right)\left(\mathrm{1}+\mathrm{tany}\right)=\mathrm{2}\:\:{if}\:\:{x}+{y}=\mathrm{45}° \\ $$$$ \\ $$
Answered by tanmay last updated on 13/May/19
$$\left(\mathrm{1}+{tanx}\right)\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right) \\ $$$$\left(\mathrm{1}+{tanx}\right)\left(\mathrm{1}+\frac{\mathrm{1}−{tanx}}{\mathrm{1}+{tanx}}\right) \\ $$$$\left(\mathrm{1}+{tanx}\right)\left(\frac{\mathrm{2}}{\mathrm{1}+{tanx}}\right) \\ $$$$\mathrm{2} \\ $$
Commented by ranusahu last updated on 14/May/19
$$\mathrm{thankyou} \\ $$
Commented by tanmay last updated on 14/May/19
$${most}\:{welcome} \\ $$
Answered by Askash last updated on 20/May/19
$$\left(\mathrm{1}+{tanx}\right)\left\{\mathrm{1}+{tan}\left(\mathrm{45}°−{x}\right)\right\} \\ $$$$\left(\mathrm{1}+{tanx}\right)\left(\frac{\mathrm{1}+{tan}\mathrm{45}°−{tanx}+{tanx}}{\mathrm{1}+{tanx}}\right) \\ $$$$\mathrm{2} \\ $$
Answered by Askash last updated on 20/May/19
$$\left(\mathrm{1}+{tanx}\right)\left\{\mathrm{1}+{tan}\left(\mathrm{45}°−{x}\right)\right\} \\ $$$$\left(\mathrm{1}+{tanx}\right)\left(\frac{\mathrm{1}+{tan}\mathrm{45}°−{tanx}+{tanx}}{\mathrm{1}+{tanx}}\right) \\ $$$$\mathrm{2} \\ $$