Question

Sin⁴A + Sin²ACos²A =Sin²A ?​

Answer 1

TO PROVE :

$\sin {}^{4} \alpha + { \sin }^{2} \alpha \cos{}^{2} \alpha = \sin{}^{2} \alpha$

PROOF :

L.H.S

$\sin {}^{2} \alpha ( \sin{}^{2} \alpha + \cos{}^{2} \alpha ) \\ \\ but \\ { \sin }^{2} \gamma + { \cos }^{2} \gamma = 1 \\ it \: is \: an \: identity \\ \\ therefore \\ sin {}^{2} \alpha ( \sin{}^{2} \alpha + \cos{}^{2} \alpha ) \\ = = > \sin {}^{2} \alpha$

L.H.S = R.H.S

HENCE PROVED

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Answer 2

By L.H.S = R,H.S

Step-by-step explanation:

Given:

Sin⁴A + Sin²ACos²A = Sin²A

Computation:

From L.H.S

⇒ Sin⁴A + Sin²ACos²A

Take Sin²A as common from the following equation;

⇒ Sin²A (Sin²A + Cos²A)

⇒ We know that; Sin²α+ Cos²α = 1

⇒ So, Sin²A + Cos²A = 1

By putting value of Sin²A + Cos²A = 1, we get

⇒ Sin²A (1)

⇒ Sin²A R.H.S

Therefore, we say that Sin⁴A + Sin²ACos²A = Sin²A

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