Question

2cot 2theta+tan theta - cot theta is equal to
​

Answer 1

the answer is 0

Given

2cot2θ + tanθ - cotθ

To Find

Simplified solution

Solution

2cot2θ + tanθ - cotθ

Applying formula tanx = sinx/cosx and cotx = cosx/sinx we get,

= $2\frac{cos2\theta}{sin2\theta} + \frac{sin\theta}{cos\theta} - \frac{cos\theta}{sin\theta}$

Now adding the last two variables up we get

$2\frac{cos2\theta}{sin2\theta} + \frac{sin^2\theta - cos^2\theta}{sin\theta cos\theta}$

Now we need to use the formula cos2x = cos^2x - sin^2x = -(sin^2x - cos^2x) to get,

$2\frac{cos2\theta}{sin2\theta} - \frac{cos2\theta}{sin\theta cos\theta}$

Now multiplying 2 in the numerator and denominator of the second variable we get

$2\frac{cos2\theta}{sin2\theta} - \frac{2cos2\theta}{2sin\theta cos\theta}$

Now, applying the formula sin2x = 2sinxcosx we get

= $2\frac{cos2\theta}{sin2\theta} - \frac{2cos2\theta}{sin2\theta}$

= $cot2\theta - cot2\theta$

= 0

Hence, the answer is 0

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