Question

(1 + sino + cos0)2 = 2(1 + sino) (1 + coso)
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Answer 1

Given Question :-

Prove that

$\sf \: {(1 + sin\theta + cos\theta)}^{2} = 2(1 + sin\theta)(1 + cos\theta)$

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

Consider LHS

$\rm :\longmapsto\: {(1 + sin\theta + cos\theta)}^{2}$

can be rewritten as

$\rm \:  =  \: {\bigg[(1 + sin\theta) + cos\theta\bigg]}^{2}$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }}}$

So, using this, we get

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

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

can be re-arranged as

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

$\rm \:  =  \:1 +1 + 2sin\theta + 2cos\theta(1 + sin\theta)$

$\rm \:  =  \:2 + 2sin\theta + 2cos\theta(1 + sin\theta)$

$\rm \:  =  \:2(1 + sin\theta) + 2cos\theta(1 + sin\theta)$

$\rm \:  =  \: 2(1 + sin\theta)(1 + cos\theta)$

Hence,

$\red{\boxed{\tt{ \sf \: {(1 + sin\theta + cos\theta)}^{2} = 2(1 + sin\theta)(1 + cos\theta)}}}$

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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