Question

lim x tends to pi/4 tan^2x-cot^2x/secx-cosecx​

Answer 1

Step-by-step explanation:

see the attachment....

#BAL

Answer 2

$\lim_{x \to \dfrac{\pi}{4} } \dfrac{\tan^2x-\cot^2x}{\sec x-\csc x​}$$=2\sqrt{2}$

Step-by-step explanation:

We have,

$\lim_{x \to \dfrac{\pi}{4} } \dfrac{\tan^2x-\cot^2x}{\sec x-\csc x​}$

To find, the value of $\lim_{x \to \dfrac{\pi}{4} } \dfrac{\tan^2x-\cot^2x}{\sec x-\csc x​} =?$

∴ $\lim_{x \to \dfrac{\pi}{4} } \dfrac{\tan^2x-\cot^2x}{\sec x-\csc x​}$

$=\lim_{x \to \dfrac{\pi}{4} } \dfrac{1+\sec^2x-(1+\csc^2x)}{\sec x-\csc x​}$

Using the trigonometric identity,

$\tan^2A=1+\sec^2A$ and $\cot^2x=1+\csc^2x$

$=\lim_{x \to \dfrac{\pi}{4} } \dfrac{1+\sec^2x-1-\csc^2x}{\sec x-\csc x​}$

$=\lim_{x \to \dfrac{\pi}{4} } \dfrac{\sec^2x-\csc^2x}{\sec x-\csc x​}$

$=\lim_{x \to \dfrac{\pi}{4} } \dfrac{(\sec x+\csc x)(\sec x-\csc x)}{\sec x-\csc x​}$

Using the identity

$a^{2} -b^{2} =(a+b)(a-b)$

$=\lim_{x \to \dfrac{\pi}{4} } \sec x+\csc x$

Put $x=\dfrac{\pi}{4}$, we get

$=\sec \dfrac{\pi}{4} +\csc \dfrac{\pi}{4}$

$=\sqrt{2} +\sqrt{2}$

$=2\sqrt{2}$

Hence, $\lim_{x \to \dfrac{\pi}{4} } \dfrac{\tan^2x-\cot^2x}{\sec x-\csc x​}$$=2\sqrt{2}$