Question

cos square theta minus sin square theta equals to cot square theta minus 1 divided by cot square theta plus 1

Answer 1

ANSWER is in picture...........

Answer 2

To prove

$\fbox{\mathsf{cos^2\theta - sin^2\theta = \frac{cot^2\theta - 1}{cot^2\theta + 1}}}$

Solving RHS

$\large\mathsf{\implies\:\frac{cot^2\theta - 1}{cot^2\theta + 1}}$

$\large\mathsf{\implies\:\frac{cot^2\theta - (cosec^2\theta - cot^2\theta)}{cot^2\theta + cosec^2\theta - cot^2\theta}}$

$\large\mathsf{\implies\:\frac{cot^2\theta - cosec^2\theta + cot^2\theta}{\cancel{cot^2\theta} + cosec^2\theta - \cancel{cot^2\theta}}}$

$\large\mathsf{\implies\: \frac{2cot^2\theta - cosec^2\theta}{cosec^2\theta}}$

$\large\mathsf{\implies\:\frac{\frac{2cos^2\theta}{sin^2\theta} - \frac{1}{sin^2\theta}}{\frac{1}{sin^2\theta}}}$

$\large\mathsf{\implies\:\frac{\frac{2cos^2\theta - 1}{\cancel{sin^2\theta}}}{\frac{1}{\cancel{sin^2\theta}}}}$

$\mathsf{\implies\:\frac{2cos^2\theta - 1}{1}}$

$\mathsf{\implies\:2cos^2\theta - 1}$

$\mathsf{\implies\:2cos^2\theta - (cos^2\theta + sin^2\theta)}$

$\mathsf{\implies\:2cos^2\theta - cos^2\theta - sin^2\theta}$

$\mathsf{\implies\:cos^2\theta - sin^2\theta}$

RHS = LHS

∴ Hence Proved

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Identity used

• cosec²θ - cot²θ = 1
• cot²θ = cos²θ / sin²θ
• cosec²θ = 1/ sin²θ
• cos²θ + sin²θ = 1