Question

If $tan\theta =\frac{1}{\sqrt{7}}$, evaluate $\frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }$ [4 MARKS]

Answer 1

Formula: 2 Marks
Proof: 2 Marks

Given $tan\theta =\frac{1}{\sqrt{7}}$
Also, the given expression is $\frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }$
Dividing the numerator and denominator by $cose{c}^2\theta$, we get
$\frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }=\frac{\frac{cose{c}^2\theta }{cose{c}^2\theta }-\frac{se{c}^2\theta }{cose{c}^2\theta }}{\frac{cose{c}^2\theta }{cose{c}^2\theta }+\frac{se{c}^2\theta }{cose{c}^2\theta }}=\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }=\frac{1-{\left(\frac{1}{\sqrt{7}}\right)}^2}{1+{\left(\frac{1}{\sqrt{7}}\right)}^2}=\frac{1-\frac{1}{7}}{1+\frac{1}{7}}=\frac{\frac{6}{7}}{\frac{8}{7}}=\frac{6}{8}=\frac{3}{4}$
Alternate Method: $tan\theta =\frac{1}{\sqrt{7}}=\frac{P}{B}\Rightarrow P=1andB=\sqrt{7}$
Now, $H^2=B^2+P^2=(\sqrt{7}{)}^2+(1{)}^2\Rightarrow H^2=8\Rightarrow H=2\sqrt{2}$
$\therefore \frac{cose{c}^2\theta -se{c}^2\theta }{cose{c}^2\theta +se{c}^2\theta }=\frac{{\left(\frac{H}{P}\right)}^2-{\left(\frac{H}{B}\right)}^2}{{\left(\frac{H}{P}\right)}^2+{\left(\frac{H}{B}\right)}^2}=\frac{{\left(\frac{2\sqrt{2}}{1}\right)}^2-{\left(\frac{2\sqrt{2}}{\sqrt{7}}\right)}^2}{{\left(\frac{2\sqrt{2}}{1}\right)}^2+{\left(\frac{2\sqrt{2}}{\sqrt{7}}\right)}^2}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}=\frac{\frac{48}{7}}{\frac{64}{7}}=\frac{48}{64}=\frac{3}{4}$