sec cosec-2sin cos prove that introduction to trigonometry

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Solution

Q.
Prove that: $\frac{\tan^{3} θ}{1 + \tan^{2} θ} + \frac{\cot^{3} θ}{1 + \cot^{2} θ} = \sec θ c o s e c θ - 2 \sin θ \cos θ$

Ans.
$LHS = \frac{\frac{\sin^{3} θ}{\cos^{3} θ}}{1 + \frac{\sin^{2} θ}{\cos^{2} θ}} + \frac{\frac{\cos^{3} θ}{\sin^{3} θ}}{1 + \frac{\cos^{2} θ}{\sin^{2} θ}} = \frac{\frac{\sin^{3} θ}{\cos^{3} θ}}{\frac{\cos^{2} θ + \sin^{2} θ}{\cos^{2} θ}} + \frac{\frac{\cos^{3} θ}{\sin^{3} θ}}{\frac{\sin^{2} θ + \cos^{2} θ}{\sin^{2} θ}}$

$= \frac{\sin^{3} θ}{\cos θ} + \frac{\cos^{3} θ}{\sin θ} = \frac{\sin^{4} θ + \cos^{4} θ}{\cos θ \sin θ} = \frac{(\sin^{2} θ + \cos^{2} θ)^{2} - 2 \sin^{2} θ \cos^{2} θ}{\cos θ \sin θ}$

$= \frac{1 - 2 \sin^{2} θ \cos^{2} θ}{\cos θ \sin θ} = \frac{1}{\cos θ \sin θ} - \frac{2 \sin^{2} θ \cos^{2} θ}{\cos θ \sin θ}$

$= \sec θ c o s e c θ - 2 \sin θ \cos θ = RHS$

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