Prove the following trigonometric identities-1-sin- 1-sin-tan- 2

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Derivative of Some Standard Functions

Prove the fol...

Question

Prove the following trigonometric identities.

$\frac{1 - \sin θ}{1 + \sin θ} = {(\sec θ - \tan θ)}^{2}$

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Solution

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\frac{(1 + sinθ)}{(1 - sinθ)}

\sqrt{\frac{\sec θ - 1}{\sec θ + 1}} + \sqrt{\frac{\sec θ + 1}{\sec θ - 1}} = 2 cosec θ

\sqrt{\frac{1 + \sin θ}{1 - \sin θ}} + \sqrt{\frac{1 - \sin θ}{1 + \sin θ}} = 2 \sec θ

\sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ

\frac{\sec θ - 1}{\sec θ + 1} = {(\frac{\sin θ}{1 + \cos θ})}^{2}

\frac{\sin θ + 1 - \cos θ}{\cos θ - 1 + \sin θ} = \frac{1 + \sin θ}{\cos θ}

Prove the following trigonometric identities:

$\frac{1 + c o s θ - s i n^{2} θ}{s i n θ (1 + c o s θ)} = c o t θ$

|\sqrt{\frac{1 - \sin θ}{1 + \sin θ}}| + |\sqrt{\frac{1 + \sin θ}{1 - \sin θ}}| = - \frac{2}{\cos θ}, where \frac{π}{2} < θ < π

\frac{1 - sin θ}{1 + sin θ} = (sec θ - tan θ)^{2}

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