Question

Proove that (1 + tan 0)2 + (1 + coto) =(seco + coseco )2​

Answer 1

Appropriate Question

Prove that

$\sf \: {(1 + tan x)}^{2} + {(1 + cotx)}^{2} = (secx + cosecx)^{2}$

Identities Used :-

$\boxed{\bf\: {sec}^{2}x = 1 + {tan}^{2}x}$

$\boxed{\bf\:{cosec}^{2}x = 1 + {cot}^{2}x}$

$\boxed{\bf\: tanx = \dfrac{sinx}{cosx}}$

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

$\boxed{\bf\:secx = \dfrac{1}{cosx}}$

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

$\boxed{\bf\: {sin}^{2}x + {cos}^{2}x = 1}$

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

Consider,

$\rm :\longmapsto\: {(1 + tan x)}^{2} + {(1 + cotx)}^{2}$

$\sf \: = \: 1 + { tan }^{2}x + 2 tan x + 1 + {cot}^{2}x + 2cotx$

$\sf \: = \: (1 + { tan }^{2}x) + 2 tan x + (1 + {cot}^{2}x)+ 2cotx$

$\sf \: = \: {sec}^{2}x + {cosec}^{2}x + 2( tanx + cotx)$

$\sf \: = \: {sec}^{2}x + {cosec}^{2}x + 2\bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx} \bigg)$

$\sf \: = \: {sec}^{2}x + {cosec}^{2}x + 2\bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{sinx \: cosx}\bigg)$

$\sf \: = \: {sec}^{2}x + {cosec}^{2}x + 2\bigg(\dfrac{1}{sinx \: cosx}\bigg)$

$\sf \: = \: {sec}^{2}x + {cosec}^{2}x + 2secx \: cosecx$

$\sf \: = \: {(secx + cosecx)}^{2}$

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

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