Question

If a = x seco + y tane , b = x tan +ysec e then x2-y2 =​

Answer 1

Appropriate Question

$\rm :\longmapsto\:a = xsec\theta + ytan\theta$

$\rm :\longmapsto\:b = xtan\theta + ysec\theta$

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

Given Trigonometric equation is

$\rm :\longmapsto\:a = xsec\theta + ytan\theta - - - - (1)$

and

$\rm :\longmapsto\:b = xtan\theta + ysec\theta - - - - (2)$

On squaring (1) and (2), we get

$\rm :\longmapsto\: {a}^{2} = {(xsec\theta + ytan\theta )}^{2}$

$\rm :\longmapsto\: {a}^{2} = {x}^{2} {sec}^{2}\theta + {y}^{2} {tan}^{2} \theta + 2xysec\theta tan\theta - - - - (3)$

Also,

$\rm :\longmapsto\: {b}^{2} = {(xtan\theta + ysec\theta )}^{2}$

$\rm :\longmapsto\: {b}^{2} = {x}^{2} {tan}^{2}\theta + {y}^{2} {sec}^{2} \theta + 2xysec\theta tan\theta - - - - (4)$

On Subtracting equation (4) from (3), we get

$\rm :\longmapsto\: {a}^{2} - {b}^{2} = {x}^{2}( {sec}^{2}\theta - {tan}^{2}\theta ) - {y}^{2}({sec}^{2}\theta - {tan}^{2}\theta ))$

$\rm :\longmapsto\: {a}^{2} - {b}^{2} = {x}^{2} - {y}^{2}$

$\bf\implies \: {x}^{2} - {y}^{2} = {a}^{2} - {b}^{2}$

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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

Appropriate Question

$\rm :\longmapsto\:a = xsec\theta + ytan\theta$

$\rm :\longmapsto\:b = xtan\theta + ysec\theta$

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

Given Trigonometric equation is

$\rm :\longmapsto\:a = xsec\theta + ytan\theta - - - - (1)$

and

$\rm :\longmapsto\:b = xtan\theta + ysec\theta - - - - (2)$

On squaring (1) and (2), we get

$\rm :\longmapsto\: {a}^{2} = {(xsec\theta + ytan\theta )}^{2}$

$\rm :\longmapsto\: {a}^{2} = {x}^{2} {sec}^{2}\theta + {y}^{2} {tan}^{2} \theta + 2xysec\theta tan\theta - - - - (3)$

Also,

$\rm :\longmapsto\: {b}^{2} = {(xtan\theta + ysec\theta )}^{2}$

$\rm :\longmapsto\: {b}^{2} = {x}^{2} {tan}^{2}\theta + {y}^{2} {sec}^{2} \theta + 2xysec\theta tan\theta - - - - (4)$

On Subtracting equation (4) from (3), we get

$\rm :\longmapsto\: {a}^{2} - {b}^{2} = {x}^{2}( {sec}^{2}\theta - {tan}^{2}\theta ) - {y}^{2}({sec}^{2}\theta - {tan}^{2}\theta ))$

$\rm :\longmapsto\: {a}^{2} - {b}^{2} = {x}^{2} - {y}^{2}$

$\bf\implies \: {x}^{2} - {y}^{2} = {a}^{2} - {b}^{2}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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