Question

( Cosec A - Cot A )^2=1- cos A/1+cos A

Answer 1

`( ~~ \csc(\theta )-\cot(\theta ) ~~ )^2=\cfrac{1-\cos(\theta )}{1+\cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ \csc(\theta )-\cot(\theta ) ~~ )^2\implies \csc^2(\theta )-2\csc(\theta )\cot(\theta )+\cot^2(\theta ) \\\\\\ \cfrac{1^2}{\sin^2(\theta )}-2\cdot \cfrac{1}{\sin(\theta )}\cdot \cfrac{\cos(\theta )}{\sin(\theta )}+\cfrac{\cos^2(\theta )}{\sin^2(\theta )}\implies \cfrac{1}{\sin^2(\theta )}-\cfrac{2\cos(\theta )}{\sin^2(\theta )}+\cfrac{\cos^2(\theta )}{\sin^2(\theta )}`

`\cfrac{\cos^2(\theta )-2\cos(\theta )+1}{\sin^2(\theta )}\implies \cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{\sin^2(\theta )} \\\\\\ \cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{1-\cos^2(\theta )}\implies \cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{-[\cos^2(\theta )-1]}`

`\cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{-[\cos^2(\theta )-1^2]}\implies \cfrac{[\cos(\theta )-1][\cos(\theta )-1]}{-[\cos(\theta )-1][\cos(\theta )+1]} \\\\\\ \cfrac{\cos(\theta )-1}{-[\cos(\theta )+1]}\implies \cfrac{-[\cos(\theta )-1]}{\cos(\theta )+1}\implies \cfrac{1-\cos(\theta )}{1+\cos(\theta )}`