Question

(viii) (1 + tan A. tan B)? + (tan A-tan B)? + B - ) sec? A sec? B​

Answer 1

Appropriate Question :-

Prove that

$\rm \: {(1 + tanA \: tanB)}^{2} + {(tanA - tanB)}^{2} = {sec}^{2}A \: {sec}^{2}B$

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

Consider LHS

$\rm \: {(1 + tanA \: tanB)}^{2} + {(tanA - tanB)}^{2}$

$\rm \: =(1 + {tan}^{2}A {tan}^{2}B + 2tanAtanB) +( {tan}^{2}A + {tan}^{2}B - 2tanAtanB)$

$\rm \: =1 + {tan}^{2}A {tan}^{2}B +{tan}^{2}A + {tan}^{2}B$

can be re-arranged as

$\rm \: =1 + {tan}^{2}A + {tan}^{2}A {tan}^{2}B + {tan}^{2}B$

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

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

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

$\rm \: =  \: {sec}^{2}A \: {sec}^{2}B$

Hence,

$\boxed{\rm{{(1 + tanA tanB)}^{2} + {(tanA - tanB)}^{2} = {sec}^{2}A \: {sec}^{2}B}}$

$\rule{190pt}{2pt}$

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