2 sin- tan- 1 - tan- - 2 sin- 2 -1 - tan-2 -

The correct option is B 2 sinθ1 + tanθ 2sinθtanθ(1 tanθ)+2sinθsec^2 θ(1+tanθ)^2=2sinθtanθ 2sinθtan^2 θ+2sinθ+2sinθtan^2 θ(1+tanθ)^2 #160; #160; ...

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Trigonometric Ratios of Allied Angles

2 sinθ tanθ 1...

Question

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

\frac{sin θ}{1 + tan θ}

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\frac{2 sin θ}{1 + tan θ}

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\frac{2 sin θ}{(1 + tan θ)^{2}}

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none of these

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

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

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

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

\frac{s i n (Θ + ϕ) - 2 s i n Θ + s i n (Θ - ϕ)}{c o s (Θ + ϕ) - 2 c o s Θ + c o s (θ - ϕ)} = t a n Θ

The expression $E = \frac{sec θ + tan θ - 1}{tan θ - sec θ + 1}$ can be simplified to -

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