If cosA-sinA=√2sinA.show that, cosecA=2√2cosA

1.	If cosA-sinA=√2sinA.show that, cosecA=2√2cosA
Answer» GIVEN :- $\rm :\longmapsto\:cosA - sinA = \sqrt{2} sinA$ To Show :- $\rm :\longmapsto\:cosecA = 2 \sqrt{2}cosA$ Formula Used :- $\boxed{ \sf \: {cosec}^{2}x - {cot}^{2}x = <klux>1</klux>}$ $\boxed{ \sf \: cotx = \dfrac{cosx}{sinx}}$ $\boxed{ \sf \: cosecx = \dfrac{1}{sinx}}$ Solution :- Given that, $\rm :\longmapsto\:cosA - sinA = \sqrt{2} sinA$ $\rm :\longmapsto\:cosA = \sqrt{2} sinA + sinA$ $\rm :\longmapsto\:cosA = ( \sqrt{2} +1) sinA$ $\rm :\longmapsto\:\dfrac{cosA}{sinA} = \sqrt{2} + 1$ $\rm :\longmapsto\:cotA = \sqrt{2} +1 - - (1)$ We know, $\rm :\longmapsto\: {cosec}^{2}A = 1 + {cot}^{2}A$ $\rm :\longmapsto\: {cosec}^{2}A = 1 + {( \sqrt{2} + 1)}^{2}$ $\rm :\longmapsto\: {cosec}^{2}A = 1 + 2 + 1 + 2 \sqrt{2}$ $\rm :\longmapsto\: {cosec}^{2}A = 4 + 2 \sqrt{2}$ $\rm :\longmapsto\:cosecA = \sqrt{4 + 2 \sqrt{2} }$ $\rm :\longmapsto\:cosecA = \sqrt{2} \sqrt{2 + \sqrt{2} } - - (2)$ Now, From equation (1), we have $\rm :\longmapsto\:cotA = \sqrt{2} +1$ $\rm :\longmapsto\: \dfrac{cosA}{sinA} = \sqrt{2} +1$ $\rm :\longmapsto\:cosA \times cosecA = \sqrt{2} + 1$ can be rewritten as, $\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times {cosec}^{2}A = \sqrt{2} + 1$ $\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times 2(2 + \sqrt{2}) = \sqrt{2} + 1$ $\: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because\:cosecA = \sqrt{2} \sqrt{2 + \sqrt{2} } }$ $\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times 2( \sqrt{2} \times \sqrt{2} + \sqrt{2}) = \sqrt{2} + 1$ $\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times 2 \sqrt{2} \: \: \cancel{( \sqrt{2}+1)} = \cancel{\sqrt{2} + 1}$ $\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times 2 \sqrt{2} = 1$ $\bf\implies \:cosecA = 2 \sqrt{2}cosA$ ${{\boxed{\bf{Hence, Proved}}}}$ 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

Answer»

GIVEN :-

$\rm :\longmapsto\:cosA - sinA = \sqrt{2} sinA$

To Show :-

$\rm :\longmapsto\:cosecA = 2 \sqrt{2}cosA$

Formula Used :-

$\boxed{ \sf \: {cosec}^{2}x - {cot}^{2}x = <klux>1</klux>}$

$\boxed{ \sf \: cotx = \dfrac{cosx}{sinx}}$

$\boxed{ \sf \: cosecx = \dfrac{1}{sinx}}$

Solution :-

Given that,

$\rm :\longmapsto\:cosA - sinA = \sqrt{2} sinA$

$\rm :\longmapsto\:cosA = \sqrt{2} sinA + sinA$

$\rm :\longmapsto\:cosA = ( \sqrt{2} +1) sinA$

$\rm :\longmapsto\:\dfrac{cosA}{sinA} = \sqrt{2} + 1$

$\rm :\longmapsto\:cotA = \sqrt{2} +1 - - (1)$

We know,

$\rm :\longmapsto\: {cosec}^{2}A = 1 + {cot}^{2}A$

$\rm :\longmapsto\: {cosec}^{2}A = 1 + {( \sqrt{2} + 1)}^{2}$

$\rm :\longmapsto\: {cosec}^{2}A = 1 + 2 + 1 + 2 \sqrt{2}$

$\rm :\longmapsto\: {cosec}^{2}A = 4 + 2 \sqrt{2}$

$\rm :\longmapsto\:cosecA = \sqrt{4 + 2 \sqrt{2} }$

$\rm :\longmapsto\:cosecA = \sqrt{2} \sqrt{2 + \sqrt{2} } - - (2)$

Now,

From equation (1), we have

$\rm :\longmapsto\:cotA = \sqrt{2} +1$

$\rm :\longmapsto\: \dfrac{cosA}{sinA} = \sqrt{2} +1$

$\rm :\longmapsto\:cosA \times cosecA = \sqrt{2} + 1$

can be rewritten as,

$\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times {cosec}^{2}A = \sqrt{2} + 1$

$\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times 2(2 + \sqrt{2}) = \sqrt{2} + 1$

$\: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because\:cosecA = \sqrt{2} \sqrt{2 + \sqrt{2} } }$

$\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times 2( \sqrt{2} \times \sqrt{2} + \sqrt{2}) = \sqrt{2} + 1$

$\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times 2 \sqrt{2} \: \: \cancel{( \sqrt{2}+1)} = \cancel{\sqrt{2} + 1}$

$\rm :\longmapsto\:\dfrac{cosA}{cosecA} \times 2 \sqrt{2} = 1$

$\bf\implies \:cosecA = 2 \sqrt{2}cosA$

${{\boxed{\bf{Hence, Proved}}}}$

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