Question

${\sec }^{2}x-cose{c}^{2}x=$

• A. 1
• B. ${\tan }^{2}x+{\cot }^{2}x$
• C. ${\cot }^{2}x-{\tan }^{2}x$
• D. ${\tan }^{2}x-{\cot }^{2}x$

Answer 1

The correct option is D ${\tan }^{2}x-{\cot }^{2}x$

When we look at the options, they are in terms of $\tan x$ and $\cot x.$ So, we will try to express ${\sec }^{2}x$ and $cose{c}^{2}x$ in terms of $\tan x$ and $\cot x$.

We know ${\sec }^{2}x=1+{\tan }^{2}x$ and

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

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