Question

If $\theta$ is an acute angle and $\sin \theta =\cos \theta$, find $3{\tan }^2\theta +2{\sin }^2\theta +{\cos }^2\theta -1$.

Answer 1

Given that $\sin \theta =\cos \theta$

Simplify the expression in terms of $\tan$.

$\sin \theta =\cos \theta$

$\frac{\sin \theta }{\cos \theta }=1$

$\tan \theta =1$

Since, $\theta$ is an acute angle. Therefore, $\theta =\frac{\pi }{4}$.

Substitute $\theta =\frac{\pi }{4}$ in $3{\tan }^2\theta +2{\sin }^2\theta +{\cos }^2\theta -1$

$3{\tan }^2\left(\frac{\pi }{4}\right)+2{\sin }^2\left(\frac{\pi }{4}\right)+{\cos }^2\left(\frac{\pi }{4}\right)-1$. (1)

Since, $\sin \left(\frac{\pi }{4}\right)=\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$ and $\tan \theta =1$.

Simplify equation (1).

$\Rightarrow 3{\left(1\right)}^2+2{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2-1$

$\Rightarrow 3+1+\frac{1}{2}-1$

$\Rightarrow \frac{7}{2}$

Thus, the value of $3{\tan }^2\theta +2{\sin }^2\theta +{\cos }^2\theta -1$ is $\frac{7}{2}$.