Question

If $\theta$ is an acute angle and $\sin \theta =\cos \theta$, find the value of $2{\tan }^2\theta +{\sin }^2\theta -1$.

Answer 1

$\sin \theta =\cos \theta$

$\frac{\sin \theta }{\cos \theta }=\frac{\cos \theta }{\cos \theta }$

$\tan \theta =1=\tan {45}^o$

$\theta ={45}^o$

Therefore, $2{\tan }^{\theta }+{\sin }^2\theta -1=2{\tan }^4{5}^o+{\sin }^2{45}^o-1$

$=2\times (1{)}^2+(\frac{1}{\sqrt{2}}{)}^2-1=2+\frac{1}{2}-1=\frac{3}{2}$.