If x = 2 ln cot t and y = tan t + cot t, the value of dy / dx isa)cot 2tb)tan 2tc)cos2td)sec2tCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

To find dy/dx, we need to differentiate y with respect to x using the chain rule.

Given:
x = 2 ln(cot(t))
y = tan(t) * cot(t)

Let's find dx/dt and dy/dt first:

dx/dt:
Using the chain rule, we have:
dx/dt = d/dt [2 ln(cot(t))]

To differentiate ln(cot(t)), we can use the chain rule again:
d/dt [ln(cot(t))] = (1/cot(t)) * (-csc^2(t)) = -csc^2(t) * cot(t)

Therefore,
dx/dt = d/dt [2 ln(cot(t))] = 2 * -csc^2(t) * cot(t) = -2csc^2(t) * cot(t)

dy/dt:
dy/dt = d/dt [tan(t) * cot(t)]

Using the product rule, we have:
dy/dt = d/dt [tan(t)] * cot(t) + tan(t) * d/dt [cot(t)]

To differentiate tan(t), we use the chain rule:
d/dt [tan(t)] = sec^2(t)

To differentiate cot(t), we can use the chain rule again:
d/dt [cot(t)] = -csc^2(t)

Therefore,
dy/dt = sec^2(t) * cot(t) + tan(t) * -csc^2(t)
= sec^2(t) * cot(t) - tan(t) * csc^2(t)

Now, let's find dy/dx:

dy/dx = (dy/dt) / (dx/dt)
= (sec^2(t) * cot(t) - tan(t) * csc^2(t)) / (-2csc^2(t) * cot(t))

Simplifying the expression:
dy/dx = (sec^2(t) * cot(t) - tan(t) * csc^2(t)) / (-2csc^2(t) * cot(t))
= -sec^2(t) / 2
= -0.5sec^2(t)

Therefore, the correct answer is option B: tan(2t).

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