Question

In the given figure, AD = DB and $\angle$B is a right angle. Determine :
${\sin }^2\theta +{\cos }^2\theta$

Answer 1

AB = a

$\Rightarrow$ AD + DB = a

$\Rightarrow$ AD + AD = a

$\Rightarrow$ 2 AD = a

$\Rightarrow$ AD = $\frac{a}{2}$

Thus, AD = DB = $\frac{a}{2}$

By Pythagoras theorem, we have

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

$\Rightarrow b^2=a^2+B{C}^2$

$\Rightarrow B{C}^2=b^2-a^2$

$\Rightarrow BC=\sqrt{b^2-a^2}$

Thus, in $\Delta$BCD, we have

Base = BC = $\sqrt{b^2-a^2}$ and Perpendicular = BD = $\frac{a}{2}$

Applying Pythagoras theorem in $\Delta$ BCD, we have

$\Rightarrow B{C}^2+B{D}^2=C{D}^2$

$\Rightarrow (\sqrt{b^2-a^2)}+{\left(\frac{a}{2}\right)}^2=C{D}^2$

$\Rightarrow C{D}^2=b^2-a^2+\frac{a^2}{4}$

$\Rightarrow C{D}^2=\frac{4b^2-4a^2+a^2}{4}$

$\Rightarrow C{D}^2=\frac{4b^2-3a^2}{4}$

$\Rightarrow CD=\frac{\sqrt{4b^2-3a^2}}{2}$

Now,

$si{n}^2\theta +co{s}^2\theta ={\left(\frac{a}{\sqrt{4b^2-3a^2}}\right)}^2+{\left(\frac{2\sqrt{b^2-a^2}}{\sqrt{4b^2-3a^2}}\right)}^2$

$\Rightarrow si{n}^2\theta +co{s}^2\theta =\frac{a^2}{4b^2-3a^2}+\frac{4(b^2-a^2)}{4b^2-3a^2}$

$\Rightarrow si{n}^2\theta +co{s}^2\theta =\frac{4b^2-3a^2}{4b^2-3a^2}=1.$