Question

prove this trigonometry​

Answer 1

Answer:

Step-by-step explanation:

Take theta=x and alpha as y for convenience,

=> cos4x/cos2y+sin4x/sin2y=1

(take LCM)

=> cos4xsin2y+sin4xcos2y/cos2ysin2y=1

=> cos4xsin2y+sin4xcos2y=cos2ysin2y

(We know that, sin2y+cos2y=1, => sin2y=1-cos2y)

=> cos4x(1-cos2y)+cos2ysin4x= cos2y(1-cos2y)

=> cos4x-cos4xcos2y+sin4xcos2y=cos2y-cos4y

(sin2x=1-cos2x)

=> cos4x-cos4xcos2y+cos2y [(1-cos2x)]^2= cos2y-cos4y

=> cos4x-cos4xcos2y+cos2y(1+cos4x-2cosx)= cos2y-cos4y

=> cos4x-cos4xcos2y+cos2y+cos4xcos2y-2cos2ycos2x=cos2y-cos4y

(cancel the common terms)

=> cos4x-2cos2ycos2x+cos4y=0

=> (cos2x-cos2y)^2=0

=> So, cos2y=cos2x......1

and similarly,

1-sin2y=1-sin2x

=> sin2y=sin2x

So,

cos2x.cos2x/cos2y+sin2x.sin2y/sin2y=1.......

Hence, proved!