Question

tan ^4 theta +tan^2theta=sec^theta-sec^2theta

Answer 1

$\large\underline{\sf{To\:prove- }}$

$\rm :\longmapsto\: {tan}^{4}\theta + {tan}^{2}\theta = {sec}^{4}\theta - {sec}^{2}\theta$

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

Consider LHS

$\rm :\longmapsto\: {tan}^{4}\theta + {tan}^{2}\theta$

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

We know,

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

So, using this identity, we get

$\rm \:  =  \:( {sec}^{2}\theta - 1)[ {sec}^{2}\theta - 1 + 1]$

$\rm \:  =  \:( {sec}^{2}\theta - 1){sec}^{2}\theta$

$\rm \:  =  \: {sec}^{4}\theta - {sec}^{2}\theta$

Hence,

$\red{\rm :\longmapsto\boxed{ \tt{ \: \: {tan}^{4}\theta + {tan}^{2}\theta = {sec}^{4}\theta - {sec}^{2}\theta }}}$

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