(2+sin^6a+cos^6a)/(
sin^2a+cos^4a)
Answers
Step-by-step explanation:
The given equation is:
2(sin^{6}A+cos^{6}A)-3(sin^{4}A+cos^{4}A)+12(sin
6
A+cos
6
A)−3(sin
4
A+cos
4
A)+1
Using a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})a
3
+b
3
=(a+b)(a
2
−ab+b
2
)
⇒2(sin^{2}A+cos^{2}A)(sin^{4}A-sin^{2}Acos^{2}B+cos^{2}A)-3(sin^{4}A+cos^{4}A)+12(sin
2
A+cos
2
A)(sin
4
A−sin
2
Acos
2
B+cos
2
A)−3(sin
4
A+cos
4
A)+1
⇒2(1)(sin^{4}A-sin^{2}Acos^{2}B+cos^{4}A)-3(sin^{4}A+cos^{4}A)+12(1)(sin
4
A−sin
2
Acos
2
B+cos
4
A)−3(sin
4
A+cos
4
A)+1
⇒2sin^{4}A-2sin^{2}Acos^{2}B+2cos^{4}A-3sin^{4}A-3cos^{4}A+12sin
4
A−2sin
2
Acos
2
B+2cos
4
A−3sin
4
A−3cos
4
A+1
⇒-sin^{4}A-2sin^{2}Acos^{2}B-cos^{4}A+1−sin
4
A−2sin
2
Acos
2
B−cos
4
A+1
⇒-(sin^{4}A+2sin^{2}Acos^{2}B+cos^{4}A)+1−(sin
4
A+2sin
2
Acos
2
B+cos
4
A)+1
⇒-(sin^{2}A+cos^{2}B)^{2}+1−(sin
2
A+cos
2
B)
2
+1
⇒-1+1=0−1+1=0 = RHS
Hence proved.