Question

If sinθ+sin²θ = 1, what is the value of cos²θ+cos⁴ θ? ​

Answer 1

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

Given that,

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

can be rewritten as

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

We know,

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

So, using this, we get

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

On squaring both sides, we get

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

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

We know,

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

So, using this identity, we get

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

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

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