Question

1 + cot theta upon 1 minus cot theta + 1 - cot theta upon 1 + cot theta equal to 2 sin square theta cos square theta

Answer 1

$\textbf{To find:}$

$\dfrac{1+cot\theta}{1-cot\theta}+\dfrac{1-cot\theta}{1+cot\theta}=\dfrac{2}{sin^2\theta-cos^2\theta}$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{1+cot\theta}{1-cot\theta}+\dfrac{1-cot\theta}{1+cot\theta}$

$=\dfrac{(1+cot\theta)^2+(1-cot\theta)^2}{(1-cot\theta)(1+cot\theta)}$

$=\dfrac{1+cot^2\theta+2\,cot\theta+1+cot^2\theta-2\,cot\theta}{1-cot^2\theta}$

$=\dfrac{1+cot^2\theta+1+cot^2\theta}{1-cot^2\theta}$

$=\dfrac{2+2\,cot^2\theta}{1-cot^2\theta}$

$=\dfrac{2(1+cot^2\theta)}{1-cot^2\theta}$

$=\dfrac{2\,cosec^2\theta}{1-\frac{cos^2\theta}{sin^2\theta}}$

$=\dfrac{2\,cosec^2\theta}{\frac{sin^2\theta-cos^2\theta}{sin^2\theta}}$

$=\dfrac{2\,cosec^2\theta\,sin^2\theta}{sin^2\theta-cos^2\theta}$

$=\dfrac{2(\frac{1}{sin^2\theta})sin^2\theta}{sin^2\theta-cos^2\theta}$

$=\dfrac{2}{sin^2\theta-cos^2\theta}$

$\textbf{Answer:}$

$\boxed{\bf\dfrac{1+cot\theta}{1-cot\theta}+\dfrac{1-cot\theta}{1+cot\theta}=\dfrac{2}{sin^2\theta-cos^2\theta}}$

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