Question

The value of $\frac{{\sec }^{2}x-cose{c}^{2}x}{{\tan }^{2}x-{\cot }^{2}x}=$
$[x\in (0,\frac{\pi }{2}),x\ne \frac{\pi }{4}]$

• A. 0.0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

In the numerator we have $\sec x$ and $cosecx$ 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,

${\sec }^{2}x=1+{\tan }^{2}x$ ... (1)

$cose{c}^{2}x=1+{\cot }^{2}x$ ... (2)

Subtracting (2) from (1), we get,
${\sec }^{2}x-cose{c}^{2}x={\tan }^{2}x-{\cot }^{2}x$

$\Rightarrow \frac{{\sec }^{2}x-cose{c}^{2}x}{{\tan }^{2}x-{\cot }^{2}x}=\frac{{\tan }^{2}x-{\cot }^{2}x}{{\tan }^{2}x-{\cot }^{2}x}=1$

[Here, since $x\in (0,\frac{\Pi }{2})$ and $x\ne \frac{\Pi }{4}$, so, ${\tan }^{2}x-{\cot }^{2}x$ won't be zero.]