Question

$△ABC$ is a right angled triangle. Find the value of the
$cose{c}^{2}A+si{n}^{2}B-co{t}^{2}A$.

• A. 0.0
• B. -1
• C. 1
• D. 2

Answer 1

The correct option is D 2

$cosecA=\frac{AC}{BC}$
$cotA=\frac{AB}{BC}$
$sinB=sin{90}^{\circ }=1$

$cose{c}^{2}A+si{n}^{2}B-co{t}^{2}A=\frac{A{C}^2}{B{C}^2}+1-\frac{A{B}^2}{B{C}^2}$
$cose{c}^{2}A+si{n}^{2}B-co{t}^{2}A=\frac{A{C}^2}{B{C}^2}-\frac{A{B}^2}{B{C}^2}+1$

$cose{c}^{2}A+si{n}^{2}B-co{t}^{2}A=\frac{A{C}^2-A{B}^2}{B{C}^2}+1$
Pythagoras Theorem in $△ABC$
$A{B}^2+B{C}^2=A{C}^2$
$B{C}^2=A{C}^2-A{B}^2$
So, $cose{c}^{2}A+si{n}^{2}B-co{t}^{2}A=\frac{B{C}^2}{B{C}^2}+1$

$cose{c}^{2}A+si{n}^{2}B-co{t}^{2}A=1+1=2$