Trig-o-no-metry!

$\sin (A)+\sin ^{ 2 } (A)=1\, \, \, \Longrightarrow \, \, \, \sin ^{ 2 } (A)=1-\sin (A) \\ \\ \\ \\ \begin{aligned} \cos ^{ 2 } (A)+\cos ^{ 4 } (A) & =[1-\sin ^{ 2 } (A)]+[1-\sin ^{ 2 } (A)]^{ 2 } \\ & =[1-(1-\sin (A))]+[1-(1-\sin (A))]^{ 2 } \\ & =\sin (A)+[\sin (A)]^{ 2 } \\ & =1 \end{aligned}$

A trigonometric identity will help: $cos^{2}(A)$ + $sin^{2}(A)$ =1

equating $sin(A)$ + $sin^{2}(A)$ =1 to above identity gives:

$cos^{2}(A)$ + $sin^{2}(A)$ = $sin(A)$ + $sin^{2}(A)$

Canceling like terms yields: $sin(A)$ = $cos^{2}(A)$

Since $sin(A)$ + $sin^{2}(A)$ =1, substituting $sin(A)$ for $cos^{2}(A)$ gives:

$cos^{2}(A)$ + $(cos^{2}(A))^{2}$ =1

which is:

$cos^{2}(A)$ + $cos^{4}(A)$ =1

$SinA = \cos ^{ 2 }{ A }$ And The Rest Follows!

$Given\quad That\\ sinA+\sin ^{ 2 }{ A } =1\\ =>\quad \sin ^{ 2 }{ A } =1-\sin { A } \quad So,\\ 1-\cos ^{ 2 }{ A } =1-\sin { A } \qquad \qquad :As\quad \sin ^{ 2 }{ A } =1-\cos ^{ 2 }{ A } :\\ or\quad \cos ^{ 2 }{ A } =\sin { A } ...........\quad (h)\\ We\quad have\quad to\quad find\\ \cos ^{ 2 }{ A } \cos ^{ 4 }{ A } \\ or\quad \cos ^{ 2 }{ A } +{ (\cos ^{ 2 }{ A) } }^{ 2 }\quad Putting\quad \cos ^{ 2 }{ A } =\sin { A } \quad from\quad (h)\\ we\quad have\\ sinA+\sin ^{ 2 }{ A } =>\quad \boxed { sinA+\sin ^{ 2 }{ A } =1 } \quad \quad \quad Given\\ \\$

Sin^2(A)+Cos^2(A)=1 , from this and given equation Cos^2(A)= Sin(A) . Substitute in the required one ultimate expression will be Sin(A)+Sin^2(A)=1

sin(A) + sin^2 (A) = 1

sin(A) = 1 - sin^2 (A)

sin (A) = cos^2 (A) ===> square both sides

sin^2 (A) = cos^4 (A)

1 - cos^2 (A) = cos^4 (A)

1 = cos^2 (A) + cos^4 (A)

First U find out sinA, sinA+sin2A=1...........(i) sinA=1-sin2A sinA=cos2a.........(ii) (cos2A=1-sin2A)

Then, cos2A+cos4A=cos2A+(cos2A)^2 =sinA+(sinA)^2 =sinA+sin2A =1 So, cos2A+cos4A=1

sinA=1-sin^2A=cos^2A sin^2A=cos^4A 1-cos^2A=cos^4A cos^2A+cos^4A=1.

sinA+sin^2A=1

SIN^2A=1-SINA

COS^2A=ROOT OVER 1-SIN^2A

=SINA

AND COS^4A=(SINA)^2

=SIN^2A

THEREFORE COS^2A+ COS^4A=1[SINCE SINA+SIN^2A=1]

sin C + sin^2 A = 1 sin A + sin^2 A = 1, we know that sin^2 A = 1 – cos^2 A sin A + (1 – cos^2 A) = 1 sin A + 1 – cos^2 A = 1 sin A – cos^2 A = 0 - cos^2 A = -sin A cos^2 A = sin A cos^2 A + cos^4 A = cos^2 A + cos^2 A * cos^2 A Substitute: sin A + sin x * sin A = sin A + sin^2 A but sin A + sin^2 A = 1 therefore: cos^2 A + cos^4 A = sin x + sin2 A = 1

1-sin^2 A = Cos^2 A= SinA so we can replace COS^4A as SIN^2A and COS^2A+SIN^2A=1

sinA=1-sin^2A=cos^2A put it in the question and ans will be 1

Cos^2 a +cos^4 a = 1- sin^2 a + (1- sin^2 a)( 1- sin^2 a) = 1- sin^2 a+[(1-(1-sin a)) (1-(1-sin a))] = 1- sin^2 a+(sin^2 a) = 1