Question

If the sin theta + cos theta = √2 theta the value of a (cos theta - sin theta) :-

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Answer 1

Appropriate Question :-

$\rm \: If \: sin\theta + cos\theta = \sqrt{2}cos\theta , \: then \: value \: of \: cos\theta - sin\theta$

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

Given that,

$\rm \: sin\theta + cos\theta = \sqrt{2}cos\theta$

$\rm \: sin\theta = \sqrt{2}cos\theta - cos\theta$

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

On rationalizing the numerator, we get

$\rm \: sin\theta = (\sqrt{2} -1) cos\theta \times \dfrac{ \sqrt{2} + 1 }{ \sqrt{2} + 1}$

We know,

$\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

So, using this identity, we get

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

$\rm \: sin\theta = \dfrac{2 - 1}{ \sqrt{2} + 1 } \: cos\theta$

$\rm \: sin\theta = \dfrac{1}{ \sqrt{2} + 1 } \: cos\theta$

$\rm \: \sqrt{2}sin\theta + sin\theta = cos\theta$

$\rm\implies \:cos\theta - sin\theta = \sqrt{2}sin\theta$

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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

$\sf\red{Given:-}$

• If the sin theta + cos theta = √2 theta the value of a (cos theta - sin theta) .

$\sf\red{Solution:-}$

$\longmapsto\rm \: {sin \: \theta \: + \: cos \theta \: = \sqrt{2} \: - > \: 1 }$

$\longmapsto\rm{cos \: \theta \: - \: sin \: \theta \: = x \: - > 2 }$

Squarring on both sides :

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

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

$\longmapsto\rm \boxed{ \rm2 \: sin \theta \: . \: cos \theta \: = 1}$

Squarring on both sides :

$\rm\longmapsto{ {cos}^{2} \theta \: + \: {sin}^{2} \theta \: - \: 2 \: cos \theta \: . \: sin \theta = {x}^{2} }$

$\longmapsto \rm{1 - (1) = {x}^{2} }$

$\longmapsto \rm{ 0 = {x}^{2} }$

$\longmapsto \rm \boxed{ \rm{x = 0}}$

So,

$\longmapsto \rm \boxed { \rm \red{cos \theta \: . \: sin \theta = 0}}$

$\rule{259mm}1mm$

@Shivam

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