4 [1 – $s i n^{2} θ$ ] [1 + $t a n^{2} θ$ ] = ___.

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Solution

The correct option is A
4

We know that (1 - $s i n^{2} θ$ = $c o s^{2} θ$ and 1 + $t a n^{2} θ$ = $s e c^{2} θ$ )

4 [1 – $s i n^{2} θ$ ] [1 + $t a n^{2} θ$ ] = $4 c o s^{2} θ \times s e c^{2} θ = 4$

Because $s e c θ = \frac{1}{c o s θ}$

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Similar questions

\frac{c o s^{2} θ + t a n^{2} θ - 1}{s i n^{2} θ} = t a n^{2} θ

Solve:
$4 [1 - s i n^{2} θ] [1 + t a n^{2} θ$ ]

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

\sqrt{1 + t a n^{2} θ} \sqrt{1 + c o t^{2} θ} \sqrt{1 - c o s^{2} θ} \sqrt{1 - s i n^{2} θ} =

Simplify :
4(1 – $s i n^{2} θ$ ) (1 + $t a n^{2} θ$ )

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