(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ 
(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ 

(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ 

(ii) ML Aggarwal Solutions for Class 10 Chapter 18 - 64

Solution:

Given,

(i) sin2 θ + cos4 θ = cos2 θ + sin4 θ

L.H.S. = sin2 θ + cos4 θ

= (1 – cos2 θ) + cos4 θ

= cos4 θ – cos2 θ + 1

= cos2 θ (cos2 θ – 1) + 1

= cos2 θ (- sin2 θ) + 1

= 1 – sin2 θ cos2 θ

Now,

R.H.S. = cos2 θ + sin4 θ

= (1 – sin2 θ) + sin4 θ

= sin4 θ – sin2 θ + 1

= sin2 θ (sin2 θ – 1) + 1

= sin2 θ (-cos2 θ) + 1

= 1 – sin2 θ cos2 θ

Hence, L.H.S. = R.H.S.

ML Aggarwal Solutions for Class 10 Chapter 18 - 65