Question

cosec theta minus cot theta whole square is equal to 1 minus cos theta by 1 + cos theta

Answer 1

$\text{consider}$

$\frac{1-cos\theta}{1+cos\theta}$

$\text{Multiply both Numerator and donominator by 1-cos\,\theta }$

$\frac{1-cos\theta}{1+cos\theta}=\frac{1-cos\theta}{1+cos\theta}\times\frac{1-cos\theta}{1-cos\theta}$

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

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

$=\frac{1}{sin^2\theta}+\frac{cos^2\theta}{sin^2\theta}-2\times\frac{1}{sin\theta}\times\frac{cos\theta}{sin\theta}$

$=cosce^2\theta+cos^2\theta-2\,cosec\theta\,cos\theta$

$\text{using}$

$\boxed{\bf(a-b)^2=a^2+b^2-2ab}$

$=(cosce\theta-cos\theta)^2$

$\implies\boxed{\bf(cosce\theta-cos\theta)^2=\frac{1-cos\theta}{1+cos\theta}}$

Find more:

If cos theta +sec theta = root 3, prove that cos cube theta +sec cube theta = 0

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Answer 2

Answer:

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