Prove that $(t a n θ + c o t θ)^{2} =$ _____________

$s e c^{3} θ + 3 c o s e c^{2} θ$

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$s e c^{3} θ + c o s e c^{3} θ$

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$s e c^{3} θ + c o s e c^{2} θ$

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$s e c^{2} θ + c o s e c^{2} θ$

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Solution

The correct option is D
$s e c^{2} θ + c o s e c^{2} θ$

Consider L.H.S

$(t a n θ + c o t θ)^{2}$

$(a + b)^{2} = a^{2} + b^{2} + 2 a b$

$= t a n^{2} θ + c o t^{2} θ + 2 t a n θ c o t θ$

$= t a n^{2} θ + c o t^{2} θ + 2 (1)$ $(∵ c o t θ = \frac{1}{t a n θ})$

$= (s e c^{2} θ - 1) + (c o s e c^{2} θ - 1) + 2$ $(∵ s e c^{2} θ - 1 = t a n^{2} θ)$

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

$= s e c^{2} θ + c o s e c^{2} θ$

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