Question

If cosx-sinx=/2sinx,prove that cosx+sinx=/2cosx
​

Answer 1

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

$\rm :\longmapsto\:cosx - sinx = \sqrt{2}sinx$

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

$\rm :\longmapsto\:cosx + sinx = \sqrt{2}cosx$

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

Given that,

$\rm :\longmapsto\:cosx - sinx = \sqrt{2}sinx$

can be rewritten as

$\rm :\longmapsto\:cosx = \sqrt{2}sinx + sinx$

$\rm :\longmapsto\:cosx = (\sqrt{2} +1) sinx$

On rationalizing the numerator on RHS, we get

$\rm :\longmapsto\:cosx = (\sqrt{2} +1) sinx \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1}$

$\rm :\longmapsto\:( \sqrt{2} - 1)cosx = ( {( \sqrt{2})}^{2} - {1}^{2})sinx$

$\rm :\longmapsto\:( \sqrt{2} - 1)cosx = ( 2 - 1)sinx$

$\rm :\longmapsto\:( \sqrt{2} - 1)cosx = sinx$

$\rm :\longmapsto\: \sqrt{2}cosx - cosx = sinx$

$\bf\implies \:\:cosx + sinx = \sqrt{2}cosx$

Hence, Proved

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