Question

2+sin^6 theta +cos^6 theta/sin^2theta+cos^4 theta

Answer 1

Answer:

Step-by-step explanation:

LHS=2(sin  

6

θ+cos  

6

θ)−3(sin  

4

θ+cos  

4

θ)+1

=2{(sin  

2

θ+cos  

2

θ)  

3

−3sin  

2

θcos  

2

θ(sin  

2

θ+cos  

2

θ)}−3(sin  

2

θ+cos  

2

θ)  

2

−2(sin  

2

θcos  

2

θ)}+1

We know, [sin²x+cos²x=1]

=2{1−3sin  

2

θcos  

2

θ}−3{1−2sin  

2

θcos  

2

θ}+1

=2−6sin  

2

θcos  

2

θ−3+6sin  

2

θcos  

2

θ+1

=0

=RHS