Question

tan³0-1/ tan 0-1 = sec²0 + tan0 Prove the following R. H S=L. H. S​

Answer 1

Appropriate Question :-

Prove that

$\rm \: \dfrac{ {tan}^{3} \theta - 1}{tan\theta - 1} = {sec}^{2}\theta + tan\theta$

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

Consider, LHS

$\rm \: \dfrac{ {tan}^{3} \theta - 1}{tan\theta - 1}$

can be rewritten as

$\rm \:= \: \dfrac{ {tan}^{3} \theta - {1}^{3} }{tan\theta - 1}$

We know,

$\boxed{\sf{  \:\rm \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) \: \: }}$

So, using this identity in numerator, we get

$\rm \: = \:\dfrac{ \cancel{(tan\theta - 1)} \: ( {tan}^{2}\theta + tan\theta + 1) }{ \cancel{tan\theta - 1}}$

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

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

We know,

$\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }}$

So, using this identity, we get

$\rm \: = \: {sec}^{2}\theta + tan\theta$

Hence,

$\rm\implies \: \boxed{\sf{  \:\: \rm \: \dfrac{ {tan}^{3} \theta - 1}{tan\theta - 1} = {sec}^{2}\theta + tan\theta \: \: \: }}$

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

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