The value of the given expression ${s i n}^{4} A + {s i n}^{2} A {c o s}^{2} A$ will be equal to :

{s i n}^{2} A + {c o s}^{2} A

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{t a n}^{2} A

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{s i n}^{2} A

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{c o s}^{2} A

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Solution

The correct option is C ${s i n}^{2} A$
${sin}^{4} A + {sin}^{2} A {cos}^{2} A$
$= {sin}^{2} A ({sin}^{2} A + {cos}^{2} A)$
$= {sin}^{2} A \times 1 = {sin}^{2} A$

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Similar questions

\frac{{sin}^{2} A}{{cos}^{2} A} + \frac{{cos}^{2} a}{{sin}^{2} A} = \frac{1}{{sin}^{2} A {cos}^{2} A} - 2

\frac{s i n^{2} A}{c o s^{2} A} + \frac{c o s^{2} A}{s i n^{2} A} = \frac{1}{s i n^{2} A c o s^{2} A} - 2

cos A + {cos}^{2} A = 1,

{sin}^{2} A + {sin}^{4} A

How many of below given statements are correct?

1. sin2A = 2sinA.cosA

2.cos2A = $s i n^{2} A - c o s^{2} A$

3.tan A = $\frac{2 t a n \frac{A}{2}}{1 - t a n^{2} \frac{A}{2}}$

4. $s i n^{2} \frac{A}{2}$ = 1 - cosA

5. $t a n^{2} A$ = $\frac{1 - c o s 2 A}{1 + c o s 2 A}$

6. $s i n^{2} A$ = $\frac{2 t a n A}{1 - t a n^{2} A}$

{cos}^{4} A - {cos}^{2} A = {sin}^{4} A - {sin}^{2} A

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