Question

Prove that (sinA+cosecA)^2+(cosA+secA)^2=7+tan^A+cot^2A

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that \:{(sinA + cosecA)}^{2} +{(cosA + secA)}^{2} =7+{tan}^{2}A +{cot}^{2}A$

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

Consider,

$\rm :\longmapsto\: {(sinA + cosecA)}^{2} + {(cosA + secA)}^{2}$

$\rm \:= \:{sin}^{2}A +{cosec}^{2}A + 2(sinA)(cosecA)+{cos}^{2}A +{sec}^{2}A +2(cosA)(secA)$

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \sf\because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy\bigg \}}$

$\rm \:= \:( {sin}^{2}A+{cos}^{2}A) + ({sec}^{2}A +{cosec}^{2}A) +2(sinA)\dfrac{1}{(sinA)} + 2(cosA)\dfrac{1}{(cosA)}$

$\: \: \: \: \: \: \red{\bigg \{ \sf \: \because \:secx = \dfrac{1}{cosx} \: \: and \: \: cosecx = \dfrac{1}{sinx}\bigg \}}$

$\rm \:= \:(1) + (1 + {tan}^{2}A +1 + {cot}^{2}A) + 2 + 2$

$\: \: \: \: \: \: \: \: \: \red{\bigg \{\because{sec}^{2}x - {tan}^{2}x = 1 \: \: \bigg \}} \\ \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:{cosec}^{2}x - {cot}^{2}x = 1 \bigg \}}$

$\rm \:= \: \:7 + {tan}^{2}A + {cot}^{2}A$

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

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