Question

In the following figure :
AD $\perp$ BC, AC $=$ 26, CD $=$ 10, BC $=$ 42, $\angle DAC$=$x$ and $\angle B$=$y$.
Find the value of :$\frac{1}{si{n}^{2}y}-\frac{1}{ta{n}^{2}y}$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 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, $\frac{1}{{\sin }^{2}y}-\frac{1}{{\tan }^{2}y}={\csc }^{2}y-{\cot }^{2}y$
= $(\frac{H}{P}{)}^2-(\frac{B}{P}{)}^2$
= $(\frac{AB}{AD}{)}^2-(\frac{BD}{AD}{)}^2$
= $(\frac{40}{24}{)}^2-(\frac{32}{24}{)}^2$
= $\frac{1600-1024}{576}$
= $1$