Question

please solve it..... it's urgent....
if sinθ + sin2θ = a and cosθ + cos2θ = b then prove that (a²+b²) (a²+b²-3) = 2b​

Answer 1

Step-by-step explanation:

Let

cosθ

1

=l and

sinθ

1

=m

The above expressions can be re-written as

ax(l)+by(m)=(a

2

−b

2

) ...(i)

ax(

m

l

2

)−by(

l

m

2

)=0 ..(ii)

Hence,

(axl)=

l

2

by(m

3

)

Substituting in (i) we get

by(m)(

l

2

m

2

+1)=(a

2

−b

2

)

bym(

l

2

m

2

+l

2

)=(a

2

−b

2

)

Now,

cosθ

1

=l and

sinθ

1

=m

Hence,

l

2

1

+

m

2

1

=1

l

2

+m

2

=l

2

m

2

Substituting we get.

by(m)(

l

2

l

2

m

2

)=(a

2

−b

2

)

by(m

3

)=(a

2

−b

2

)

sinθ=(

a

2

−b

2

by

)

1/3

...(iii)

Similarly we can get

cosθ=(

a

2

−b

2

ax

)

1/3

...(iv)

Squaring and adding (iii) and (iv) we get

(

a

2

−b

2

by

)

2/3

+(

a

2

−b

2

ax

)

2/3

=1

(by)

2/3

+(ax)

2/3

=(a

2

−b

2

)

2/3