Question

The value of sin⁶θ + cos⁶θ+3 sin²θcos²θ is

a.1

b.3

c.0

d.-1

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Answer 1

Hey mate!!!

Here is the answer to your query...

sin^6A + cos^6A + 3sin^2Acos^2A  = (sin^2)^3 + (cos^2)^3 + 3sin^2Acos^2A

[a^3 + b^3 = (a+b)(a^2 + b^2 - ab)]

=(sin^2 + cos^2)(sin^4 + cos^4 - sin^2 cos^2) + 3sin^2cos^2

(sin^2 + cos^2 = 1)

=(1)(sin^4 + cos^4 - sin^2cos^2) + 3sin^2 cos^2

= sin^4 + cos^4 - sin^2cos^2 + 3sin^2 cos^2

= sin^4 + cos^4 + 2sin^2cos^2

[a^2 + b^2 + 2ab = (a+b)^2]

= (sin^2 + cos^2)^2

=(1)^2

= 1

Hence, option a is correct.

Hope it helps!!!

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