Question

If $\sin A-\cos A=0$, find the value of $2{\tan }^{2}A+{\sin }^{2}A+1$.

Answer 1

$\begin{array}{c}As\sin A=\cos A\\ \frac{p}{h}=\frac{b}{h}\\ perpendicular=base\\ letthepbe1\\ \therefore perpendicular=1\\ base=1\\ hypotenuse=\sqrt{2}\left(usingpythagorastheoram\right)\\ \sin A=\frac{p}{h}=\frac{1}{\sqrt{2}}\\ \tan A=\frac{p}{b}=\frac{1}{1}\\ \therefore 2{\tan }^{2}A+{\sin }^{2}A+1\\ =2{\left(\frac{1}{1}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2+1\\ =2+\frac{1}{2}+1\\ =\frac{4+1+2}{2}\\ =\frac{7}{2}\end{array}$

Hence, this is the answer.