Question

1 + 3 cos square theta into cot square theta + cot cot 6 theta is equal to cos 6 theta

Answer 1

SOLUTION

TO PROVE

$\sf{1 + 3 \: { cosec}^{2} \theta \: {cot}^{2} \theta + {cot}^{6} \theta = {cosec}^{6} \theta }$

EVALUATION

We are aware of the Trigonometric formula that

$\sf{{ cosec}^{2} \theta - \: {cot}^{2} \theta = 1 }$

Cubing both sides we get

$\sf{({{ cosec}^{2} \theta - \: {cot}^{2} \theta )}^{3} = 1 }$

$\sf{ \implies \: ({{ cosec}^{2} \theta) }^{3}-{ (\: {cot}^{2} \theta )}^{3} -3{ cosec}^{2} \theta . \: {cot}^{2} \theta ({ cosec}^{2} \theta - \: {cot}^{2} \theta )= 1 }$

$\sf{ \implies \: {cosec}^{6} \theta - {cot}^{6} \theta - 3 \: { cosec}^{2} \theta \: {cot}^{2} \theta .1 - 1 = 0 }$

$\sf{ \implies \: {cosec}^{6} \theta = 1 + 3 \: { cosec}^{2} \theta \: {cot}^{2} \theta + {cot}^{6} \theta }$

$\sf{ \implies \: 1 + 3 \: { cosec}^{2} \theta \: {cot}^{2} \theta + {cot}^{6} \theta = {cosec}^{6} \theta }$

Hence proved

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Answer 2

$\large\underline{\sf{Appropriate \:Question - }}$

Prove that

$\rm \: 1 + 3 {cosec}^{2}\theta {cot}^{2} \theta + {cot}^{6}\theta = {cosec}^{6}\theta$

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

Consider,

$\rm :\longmapsto\: {cosec}^{6}\theta$

can be rewritten as

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

We know that,

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

So, using this

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

We know that

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

So, using this identity, we get

$\rm \:  =  \:  \: {(1)}^{3} + {( {cot}^{2} \theta )}^{3} + 3.1. {cot}^{2}\theta (1 + {cot}^{2}\theta )$

$\rm \:  =  \:  \: 1 + {cot}^{6}\theta + 3 {cot}^{2}\theta {cosec}^{2} \theta$

$\red{\bigg \{ \because \: {cosec}^{2}x - {cot}^{2}x = 1\bigg \}}$

Hence,

$\boxed{ \bf{ \: \bf \: 1 + 3 {cosec}^{2}\theta \: {cot}^{2} \theta + {cot}^{6}\theta = {cosec}^{6}\theta }}$

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