Prove that 2sin-tan- 1-tan- -2sin- 2 - 1-tan- 2 - 2sin- 1-tan-

L.H.S. = 2sinθ[ tanθ tan ^ 2 θ + sec ^ 2 θ #160; ] #160; ( 1+tanθ #160; ) #160; ^ 2 = 2sinθ( tanθ +1 ) #160; ( 1+tanθ #160; ) #160; ^ ...

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Basic Differentiation Rule

Prove that ...

Question

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

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

\frac{tan θ + sec θ - 1}{tan θ - sec θ - 1} = sec θ + 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{1}{sec θ - tan θ} = sec θ + tan θ

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