Question

In the given figure : $AD\perp BC$, $AC=26$,
$CD=10$, $BC=42$,
$\angle DAC=x$ and
$\angle B=y$.

Find the value of :$\frac{1}{{\sin }^{2}y}-\frac{1}{{\tan }^{2}y}$

Answer 1

Given: $△ABC$, $AD\perp BC$, $AC=26$, $CD=10$, $BC=42$

Now, In $△ACD$,

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

${26}^2={10}^2+A{D}^2$

$A{D}^2=576$

$AD=24$

Now, In $△ABD$,

$BD=AC-CD$

$BD=42-10$

$BD=32$

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

$A{B}^2={32}^2+{24}^2$

$A{B}^2=1600$

$AB=40$

Now, $\frac{1}{{\sin }^{2}y}-\frac{1}{{\tan }^{2}y}={\csc }^{2}y-{\cot }^{2}y$

$=$ ${\left(\frac{H}{P}\right)}^2-{\left(\frac{B}{P}\right)}^2$

$=$ ${\left(\frac{AB}{AD}\right)}^2-{\left(\frac{BD}{AD}\right)}^2$

$=$ ${\left(\frac{40}{24}\right)}^2-{\left(\frac{32}{24}\right)}^2$

$=$ $\frac{1600-1024}{576}$

$=$ $1$