Proove- 1 - sin -1 - sin - - sec - - tan -2

→ (1 sinθ/1+sinθ×1+sinθ/1 sinθ) = (1 sinθ )^2/1 sin^2θ = (1 sinθ )^2/cos^2θ = (secθ tanθ )^2 ...

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Trigonometric Ratios of Allied Angles

Proove: 1 - ...

Question

Proove:
$\frac{1 - s i n θ}{1 + s i n θ} = (s e c θ - t a n θ)^{2}$

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Solution

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

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

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

\sqrt{\frac{1 - s i n θ}{1 + s i n θ}} = s e c θ - t a n θ

Prove the following trigonometric identities:

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

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