Question
If $3\cot\theta=4,$ show that $\frac{(1-\tan^2\theta)}{(1+\tan^2\theta)}=\big(\text{cos}^2\theta-\sin^2\theta\big).$
✓
Answer
$3\cot\theta=4\Rightarrow\cot\theta=\frac43\Rightarrow\tan\theta=\frac34$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\cot\theta=\frac{\text{Base}}{\text{Perpendicular}}=\frac{\text{AB}}{\text{BC}}=\frac43$
Let AB = 4 and BC = 3
Then, by pythagoras theoram,
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
$=4^3+3^2=16+9=25$
$\Rightarrow\text{AC} = 5$
Now,
$\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac35$
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac45$
$\therefore\text{L.H.S.}=\frac{\big(1-\tan^2\theta\big)}{\big(1+\tan^2\theta\big)}$
$=\frac{1-\big(\frac{3}{4}\big)^2}{1+\big(\frac{3}{4}\big)^2}$
$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{\frac{8}{16}}{\frac{25}{16}}=\frac{8}{25}$
$\text{R.H.S.}=\cos^2\theta-\sin^2\theta=\Big(\frac45\Big)^2=\Big(\frac{3}{5}\Big)^2$
$=\frac{16}{25}-\frac{9}{25}=\frac{8}{25}$
Hence, $\frac{(1-\tan^2\theta)}{(1+\tan^2\theta)}=\big(\text{cos}^2\theta-\sin^2\theta\big).$
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