Question

tan6thta+3tan2thta×sec2thta+1=?

Answer 1

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

Given expression is

$\rm :\longmapsto\: {tan}^{6}\theta + 3 {tan}^{2}\theta {sec}^{2}\theta + 1$

We know,

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

So, using this, above can be rewritten as

$\rm \:  =  \: {tan}^{6}\theta +3 {tan}^{2}\theta ( {tan }^{2}\theta + 1) + 1$

$\rm \:  =  \: {tan}^{6}\theta + 3 {tan}^{4}\theta + 3 {tan}^{2}\theta + 1$

can be rewritten as

$\rm \:  =  \: {( {tan}^{2}\theta ) }^{3} + 3 {( {tan}^{2}\theta ) }^{2} \times 1 + 3 {tan}^{2}\theta \times {(1)}^{2} + {(1)}^{3}$

We know,

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

So, using this identity, we get

$\rm \:  =  \: {( {tan}^{2} \theta + 1)}^{3}$

$\rm \:  =  \: {( {sec}^{2} \theta)}^{3}$

$\rm \:  =  \: {sec}^{6}\theta$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: {tan}^{6}\theta + 3 {tan}^{2}\theta {sec}^{2}\theta + 1 = {sec}^{6}\theta \: }}$

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

Answer 2

Answer:

tan6thta+3tan2thta×sec2thta+1=sec6thta