Question

Prove that , sin^2 theta + cos^2 theta = 1 - 3 sin^2 theta × cos^2 theta.​

Answer 1

Appropriate Question

Prove that

$\rm :\longmapsto\: {sin}^{6}\theta + {cos}^{6}\theta = 1 - 3 {sin}^{2}\theta \: {cos}^{2}\theta$

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

Consider LHS

$\rm :\longmapsto\: {sin}^{6}\theta + {cos}^{6}\theta =$

$\rm \:  =  \: {( {sin}^{2} \theta)}^{3} + {( {cos}^{2} \theta)}^{3}$

We know

$\boxed{ \tt{ \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \: }}$

So, using this identity, we get

$\rm \:  =  \: {\bigg[ {sin}^{2}\theta + {cos}^{2}\theta\bigg]}^{3} - 3 {sin}^{2}\theta {cos}^{2}\theta( {sin}^{2}\theta + {cos}^{2}\theta)$

We know,

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

So, using this identity, we get

$\rm \:  =  \: {(1)}^{3} - 3 {sin}^{2}\theta \: {cos}^{2}\theta(1)$

$\rm \:  =  \: 1 - 3 {sin}^{2}\theta \: {cos}^{2}\theta$

Hence,

$\rm \implies\:\boxed{ \tt{ \: {sin}^{6}\theta + {cos}^{6}\theta \: =  \: 1 - 3 {sin}^{2}\theta \: {cos}^{2}\theta \: }}$

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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

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