Differentiation Question 233
Question: $ \frac{d}{dx}( \frac{\sec x+\tan x}{\sec x-\tan x} )= $
[DSSE 1979, 81; CBSE 1981]
Options:
A) $ \frac{2\cos x}{{{(1-\sin x)}^{2}}} $
B) $ \frac{\cos x}{{{(1-\sin x)}^{2}}} $
C) $ \frac{2\cos x}{1-\sin x} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{d}{dx}( \frac{\sec x+\tan x}{\sec x-\tan x} )=\frac{d}{dx}( \frac{1+\sin x}{1-\sin x} ) $ .
BETA
