Question

A right angled $\Delta ABC$ right angled at B, and $AB=\frac{b}{a},BC=\frac{a}{b},\angle C=\theta$ then $a\sin \theta .b\cos \theta$ is:

• A. $\frac{ab}{\sqrt{a^2+b^2}}$
• B. $\frac{a^2+b^2}{ab}$
• C. $\frac{a^2{b}^2}{\sqrt{a^2+b^2}}$
• D. $\frac{a^3{b}^3}{a^4+b^4}$

Answer 1

The correct option is D $\frac{a^3{b}^3}{a^4+b^4}$

In right angled $△ABC,$

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

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

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

$\Rightarrow AC=\sqrt{\frac{b^4+a^4}{a^2{b}^2}}=\frac{1}{ab}\sqrt{b^4+a^4}$

$\Rightarrow AC=\frac{\sqrt{a^4+b^4}}{ab}$

$\Rightarrow \sin \theta =\frac{AB}{AC}=\frac{\frac{b}{a}}{\frac{\sqrt{a^4+b^4}}{ab}}=\frac{b}{a}.\frac{ab}{\sqrt{a^4+b^4}}=\frac{b^2}{\sqrt{a^4+b^4}}$

$\&\cos \theta =\frac{BC}{AC}=\frac{\frac{a}{b}}{\frac{\sqrt{a^4+b^4}}{ab}}=\frac{a}{b}.\frac{ab}{\sqrt{a^4+b^4}}=\frac{a^2}{\sqrt{a^4+b^4}}$

$\therefore a\sin \theta .b\cos \theta =\left(\frac{{ab}^2}{\sqrt{a^4+b^4}}\right)\left(\frac{{ba}^2}{\sqrt{a^4+b^4}}\right)=\frac{a^3{b}^3}{(a^4+b^4)}$

Hence, the answer is $\frac{a^3{b}^3}{(a^4+b^4)}.$