Question

\\\frac{1-\sec\theta}{\tan\theta} - \frac{\tan\theta}{1-\sec\theta} = 2\cot\theta\\

Explanation:

Find a common denominator for the left-hand side

$$ \frac{1-\sec\theta}{\tan\theta} - \frac{\tan\theta}{1-\sec\theta} = \frac{(1-\sec\theta)^2 - \tan^2\theta}{\tan\theta(1-\sec\theta)} $$

Expand and simplify the numerator

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Cancel common terms and simplify to the final expression

$$ \frac{2(1-\sec\theta)}{\tan\theta(1-\sec\theta)} = \frac{2}{\tan\theta} = 2\cot\theta $$

Answer:

To verify the identity:

$$ \frac{1-\sec\theta}{\tan\theta} - \frac{\tan\theta}{1-\sec\theta} = 2\cot\theta $$

We simplify the left-hand side (LHS):

1. Combine the fractions using a common denominator:

$$ \text{LHS} = \frac{(1-\sec\theta)^2 - \tan^2\theta}{\tan\theta(1-\sec\theta)} $$

1. Expand the numerator:

$$ (1-\sec\theta)^2 - \tan^2\theta = 1 - 2\sec\theta + \sec^2\theta - \tan^2\theta $$

1. Apply the Pythagorean identity \(\sec^2\theta - \tan^2\theta = 1\):

$$ 1 - 2\sec\theta + 1 = 2 - 2\sec\theta = 2(1-\sec\theta) $$

1. Substitute back into the fraction and simplify:

$$ \text{LHS} = \frac{2(1-\sec\theta)}{\tan\theta(1-\sec\theta)} = \frac{2}{\tan\theta} = 2\cot\theta $$

Since the LHS simplifies to \(2\cot\theta\), the identity is verified.