Question

Find the value of $\frac{{\sec }^{2}x-{\text{cosec}}^{2}x}{{\tan }^{2}x-{\cot }^{2}x}$,
$(x\in (0,\frac{\pi }{2}),x\ne \frac{\pi }{4})$

Answer 1

In the numerator we have $\sec x$ and $\text{cosec}x$ and in the denominator we have $\tan x$ and $\cot x$. So we will try to express numerator in terms of denominator. Using the identities,

${\sec }^{2}x=1+{\tan }^{2}x$ $\cdot \cdot \cdot \left(1\right)$
${\text{cosec}}^{2}x=1+{\cot }^{2}x$ $\cdot \cdot \cdot \left(2\right)$
Subtracting $\left(2\right)$ from $\left(1\right)$, we get
${\sec }^{2}x-{\text{cosec}}^{2}x={\tan }^{2}x-{\cot }^{2}x$
$\Rightarrow \frac{{\sec }^{2}x-{\text{cosec}}^{2}x}{{\tan }^{2}x-{\cot }^{2}x}=\frac{{\tan }^{2}x-{\cot }^{2}x}{{\tan }^{2}x-{\cot }^{2}x}=1$

Since $x\in (0,\frac{\pi }{2})$ and $x\ne \frac{\pi }{4}$, Thus
${\tan }^{2}x-{\cot }^{2}x$ won't be zero.