A) \[2\text{ }sec\text{ }2x\]
B) \[2\cos ec2x\]
C) \[\sec 2x\]
D) \[\cos ec2x\]
Correct Answer: B
Solution :
Let \[y=\log \,(\tan x)\] On differentiating both sides, w.r.t.\[x,\] we get \[\frac{dy}{dx}=\frac{1}{\tan x}\frac{d}{dx}(\tan x)\] \[=\frac{1}{\tan x}{{\sec }^{2}}x=\frac{1}{{{\cos }^{2}}x\frac{\sin x}{\cos x}}\] \[=\frac{2}{\sin 2x}=2\cos ec2x\]
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