Question
In $\tan \theta = 1$, find the value of $5\cot^2\theta + \sin^2\theta - 1$.
✓
Answer
Consider $\triangle A B C$, where $\angle A=90^{\circ}$
$\tan \theta=\frac{\text { Perpendicular }}{\text { Basse }}=\frac{ AB }{ AC }=1=\frac{1}{1}$
By Pythagoras theorem,
$B C^2$
$=A B^2+A C^2$
$=1^2+1^2$
$=2$
$\Rightarrow B C=\sqrt{2} $
Now,
$ \cot \theta=\frac{1}{\tan \theta}=1$
$\sin \theta=\frac{ AB }{ BC }=\frac{1}{\sqrt{2}}$
$\therefore 5 \cot ^2 \theta+\sin ^2 \theta-1$
$=5 \times(1)^2+\left(\frac{1}{\sqrt{2}}\right)^2-1$
$=5+\frac{1}{2}-1$
$=4+\frac{1}{2}$
$=\frac{9}{2} . $
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