Show that -tan-2-1-sin-1-sin-

(θ tanθ)^2 =(1cosθ sinθcosθ)^2 =(1 sinθ)^2cos^2θ since cos^2θ=1 sin^2θ =1 sinθ1+sinθ Hence proved ...

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General Solution of Trigonometric Equation

Show that θ-...

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

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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}

\frac{(1 + sinθ)}{(1 - sinθ)}

Prove the following trigonometric identities:

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

Prove the following

1. $(1 - {sin}^{2} A) s e c^{2} A = 1$

2. ${sec}^{4} θ - {sec}^{2} θ = {tan}^{4} θ + {tan}^{2}$

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

4. $\frac{tan θ + sec θ - 1}{tan θ - sec θ +} = \frac{1 + sin θ}{cos θ}$

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