Question

(+ tan A tan B)² + (tann-lan B)?
Seca A Sec² B​

Answer 1

Appropriate Question :-

Prove that

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

Answer :-

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

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

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

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

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

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

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

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

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

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