Question

In a $△ABC$, in which $\angle ={90}^o,\angle ABC={\theta }^o,BC=21$ units, $AB=29$ units, show that $(co{s}^2\theta -si{n}^2\theta )=\frac{41}{841}$

Answer 1

Draw a triangle using given instructions:
From figure : $△ABC$ is a right angled triangle
$A{B}^2=A{C}^2+B{C}^2$
$(29{)}^2=A{C}^2+(21{)}^2$
$841=A{C}^2+441$
$A{C}^2=400$
or $AC=20$

Find $\sin \theta$ and $\cos \theta$
$sin\theta =\frac{AC}{AB}=\frac{20}{29}$
$cos\theta =\frac{BC}{AB}=\frac{21}{29}$

Now :
LHS = $co{s}^2\theta -si{n}^2\theta$
$={\left(\frac{21}{29}\right)}^2-{\left(\frac{20}{29}\right)}^2$
$=\frac{41}{841}$
= RHS
Hence proved.