Question

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

Answer 1

In $△ABC,\frac{BC}{AC}=\frac{24}{25}$

Applying Pythagoras Theorem:

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

$\Rightarrow A{B}^2=A{C}^2-B{C}^2$

$\Rightarrow A{B}^2={25}^2-{24}^2=(25-24)(25+24)=49$

$\Rightarrow AB=7$

Now, $cosecA=\frac{AC}{BC}=\frac{25}{24}$ and $\cot A=\frac{AB}{BC}=\frac{7}{24}$
$\Rightarrow cose{c}^{2}A–{\cot }^{2}A=\frac{{25}^2-7^2}{{24}^2}=\frac{576}{576}=1$

$\Rightarrow cose{c}^{2}A–{\cot }^{2}A=1$, which is also an identity.