Question

answer this question fast........please don't give irrelevant answers
​

Answer 1

Cos^6A+sin^6A

(Cos3A4)2+(Sin3A)2

(Cos2A+Sin2A)(Cos4A+sin4A-cos2A×sin2A)

{A3+b3=(a+b)(A2+B2-ab)

(1)(cos2A2)+(sin2A)2-2cos2A×sin2A-Cos2A×sin2A

{A2+B2={(a+b)2-2ab}

12-3sin2A×cos2A

1-3sin2A×cos2A. RHS proved

according to above answer the number are squares

ex:b3 is 3 is square of b

Answer 2

Answer:

Step-by-step explanation:

cos^4α+2cos^2α(1-cos^2α),                                 $\bold{\frac{1}{sec \alpha} =cos \alpha}$.

cos^4α+2cos^2α-2cos^4α,

2 cos^2α-cos^4α,

cos^2 α(2-cos^2 α),

(1-sin^2 α)(2-(1-sin^2 α)),

(1-sin^2 α)(1+ sin^2 α),

(1)^2-(sin^2 α)^2=1 ,

So, 1-sin^4 α is the answer.

I will give irrelevant answer kya karogii bevakoof hahahhaa:)