Question

sec theta minus tan theta the whole square equal to 1 minus sin theta by 1 + sin theta prove

Answer 1

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Answer 2

Answer:

To Prove: $(sec\,\theta-tan\,\theta)^2=\frac{1-sin\,\theta}{1+sin\,\theta}$

Consider,

LHS

$=(sec\,\theta-tan\,\theta)^2$

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

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

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

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

$=\frac{(1-sin\,\theta)^2}{(1-sin\,\theta)(1+sin\,\theta)}$

$=\frac{1-sin\,\theta}{1+sin\,\theta}$

$=RHS$

Hence Proved