Question

If tan $\theta =\frac{1}{\sqrt{7}}$, show that $\frac{cose{c}^2\theta -se{c}^2\theta }{(cose{c}^2\theta +se{c}^2\theta }=\frac{3}{4}$

Answer 1

Let us consider a right $\Delta ABC$ right angled at B

Now it is given that $\tan \theta =\frac{AB}{BC}=\frac{1}{\sqrt{7}}$

So, if AB = k, then BC = $\sqrt{7}k$ where is a positive number

Using Pythaforas theorem, we have

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=(k{)}^2+(\sqrt{7}k{)}^2$

$A{C}^2=k^2+7k^2=8k^2$

$AC=\sqrt{8}k=2\sqrt{2}k$

Now

$sin\theta =\frac{AB}{AC}=\frac{k}{2\sqrt{2}k}=\frac{1}{2\sqrt{2}}$

$cos\theta =\frac{BC}{AC}=\frac{\sqrt{7}k}{2\sqrt{2}k}=\frac{\sqrt{7}}{2\sqrt{2}}$

$sec\theta =\frac{2\sqrt{2}}{\sqrt{7}}$

$cosec\theta =2\sqrt{2}$

Substitute these values in

$\frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }$

$=\frac{(2\sqrt{2}{)}^2-(\frac{2\sqrt{2}}{\sqrt{7}}{)}^2}{(2\sqrt{2}{)}^2+(\frac{2\sqrt{2}}{\sqrt{7}}{)}^2}$

$=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$

$=\frac{\frac{56-8}{7}}{\frac{56+8}{7}}$

$=\frac{48}{64}=\frac{3}{4}$

Hence proved