Question

In the given figure, $\frac{BC}{AC}$ = $\frac{24}{25}$ .Find the value of $cose{c}^{2}A$ – $co{t}^{2}A$ .

Answer 1

cosec A = = $\frac{AC}{BC}$ = $\frac{25}{24}$

Using Pythagoras theorem , we have $A{C}^2=A{B}^2+B{C}^2$

Let $BCbe24xandACbe25x$

$(25x{)}^2=A{B}^2+(24x{)}^2$

$A{B}^2=625x^2-576x^2$

= $49x^2$

$AB=7x$

cot A = $\frac{AB}{BC}$

cot A = $\frac{7x}{24x}$
cot A = $\frac{7}{24}$

$cose{c}^{2}A$ – $co{t}^{2}A$ = 1