Question

cos theta + cos^2 theta = 1 then find sin^2 theta + sin^4 theta

Answer 1

$\green{\large\underline{\sf{Given- }}}$

$\boxed{ \tt{ \: cos\theta + {cos}^{2}\theta = 1}}$

$\blue{\large\underline{\sf{To\:Find - }}}$

$\boxed{ \tt{ \: {sin}^{2}\theta + {sin}^{4} \theta }}$

$\red{\large\underline{\sf{Solution-}}}$

Given that

$\rm :\longmapsto\: \: cos\theta + {cos}^{2}\theta = 1$

can be rewritten as

$\rm :\longmapsto\:cos\theta = 1 - {cos}^{2}\theta$

We know,

$\boxed{ \tt{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

So, using this, we get

$\rm :\longmapsto\:cos\theta = {sin}^{2}\theta$

On squaring both sides, we get

$\rm :\longmapsto\: {cos}^{2}\theta = {( {sin}^{2} \theta )}^{2}$

$\rm :\longmapsto\:1 - {sin}^{2}\theta = {sin}^{4}\theta$

$\bf\implies \: {sin}^{2}\theta + {sin}^{4} \theta = 1$

Hence,

$\bf\implies \:\boxed{ \tt{ \: {sin}^{2}\theta + {sin}^{4} \theta = 1 \: }}$

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Answer 2

GIVEN :-

$\cos \theta \: + \cos^{2} \theta = 1$

WE HAVE TO FIND :-

$\sin^{2} \theta \: + \: \sin^{4} \theta$

SOLUTION :-

Given that

$\rightarrow \: \cos \theta \: + \cos^{2} \theta = 1$

We can write it as

$\rightarrow \cos \theta \: = 1 - \cos^{2} \theta$

We know that,

$\sf \sin^{2}x + \cos^{2} x = 1$

So, by using this, we get

$\rightarrow \cos \theta = \sin^{2} \theta$

On Squaring both sides, we get

$\rightarrow \: \cos^{2} \theta = {(\sin ^{2} \theta)}^{2} \\ \\ \rightarrow \: 1 - \sin^{2} \theta \: = \sin^{4} \theta \\ \\ \rightarrow \: \sin^{2} \theta + \sin^{4} \theta \: = 1 \\ \\ \sf \: Hence, \\ \sin^{2} \theta \: + \sin^{4} \theta = 1$