Question

If sinθ + cosθ = √2 cosθ, (θ ≠ 90°) then the value of tanθ is?​

Answer 1

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

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

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

$\rm \: tan\theta$

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

Given that,

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

can be further rewritten as

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

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

can be further rewritten as

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

We know that,

$\boxed{\sf{ \: \: \: tanx \: = \: \frac{sinx}{cosx} \: \: }}$

So, using this, we get

$\rm\implies \:tan\theta = \sqrt{2} - 1$

Hence,

$\rm\implies \: \: \boxed{\sf{ \: \: \: tan\theta = \sqrt{2} - 1 \: \: }} \:$

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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

Answer:

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Step-by-step explanation:

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