Question:
If $\tan \theta=\frac{3}{4}$, then $\cos ^{2} \theta-\sin ^{2} \theta=$
(a) $\frac{7}{25}$
(b) 1
(C) $\frac{-7}{25}$
(d) $\frac{4}{25}$
Solution:
Given that: $\tan \theta=\frac{3}{4}$
Since $\tan x=\frac{\text { Perpendicular }}{\text { Base }}$
$\Rightarrow$ Perpendicular $=3$
$\Rightarrow$ Base $=4$
$\Rightarrow$ Hypotenuse $=\sqrt{9+16}$
$\Rightarrow$ Hypotenuse $=5$
We know that $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ and $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$
We find:
$\cos ^{2} \theta-\sin ^{2} \theta$
$(4)^{2}(3)^{2}$
$=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}$
$=\frac{16}{25}-\frac{9}{25}$
$=\frac{7}{25}$
Hence the correct option is (a)