Question

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Answer 1

Answer:

2(sin6θ + cos6θ) - 3 (sin4θ + cos4θ) + 1  

=2[(sin2θ)3 + (cos2θ)3] -3[(sin2θ)2 + (cos2θ)2]+1

=2[(sin2θ + cos2θ)3 -3sin2θcos2θ (sin2θ + cos2θ)] -3[(sin2θ + cos2θ)2

-2sin2θcos2θ]+1

The algebraic identity

a3 + b3 = (a+b)3 - 3ab(a+b) and  

a2 + b2 = (a+b)2 - 2ab

are used in the above step where

a = sin2θ and b = cos2θ.

writing sin2θ + cos2θ = 1, we have  

= 2[1-3 sin2θ cos2θ] -3[-2 sin2θcos2θ] + 1

= 2-6 sin2θcos2θ -3 + 6 sin2θcos2θ + 1

= -3+3=0              

HOPE U LIKE IT PLZ MARK IT AS THE BRAINLIEST

Answer 2

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

Prove the following :

$\rm :\longmapsto\:2( {sin}^{6}\theta + {cos}^{6}\theta ) + 1 = 3( {sin}^{4}\theta + {cos}^{4}\theta )$

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

$\rm :\longmapsto\:2( {sin}^{6}\theta + {cos}^{6}\theta ) + 1$

can be rewritten as

$\rm \: = \: 2\bigg( {( {sin}^{2} \theta )}^{3} + {( {cos}^{2} \theta )}^{3} \bigg) + 1$

We know,

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

So, using this identity, we get

$\rm \: = \: 2\bigg( {( {sin}^{2}\theta + {cos}^{2} \theta ) }^{3} - 3 {sin}^{2} \theta {cos}^{2} \theta ( {sin}^{2} \theta + {cos}^{2} \theta ) \bigg) + 1$

We know,

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

So, using this identity, we get

$\rm \: = \: 2\bigg( {( 1 ) }^{3} - 3 {sin}^{2} \theta {cos}^{2} \theta ( 1 ) \bigg) + 1$

$\rm \: = \: 2\bigg( 1 - 3 {sin}^{2} \theta {cos}^{2} \theta \bigg) + 1$

$\rm \: = \: 2 - 6{sin}^{2} \theta {cos}^{2} \theta + 1$

$\rm \: = \: 3 - 6{sin}^{2} \theta {cos}^{2} \theta$

$\rm \: = \: 3(1 - 2{sin}^{2} \theta {cos}^{2} \theta)$

$\rm \: = \: 3 \bigg( {(1)}^{2} - 2{sin}^{2} \theta {cos}^{2} \theta \bigg)$

We know,

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

So, Replace 1 by this identity, we get

$\rm \: = \: 3 \bigg( {( {sin}^{2}\theta + {cos}^{2}\theta )}^{2} - 2{sin}^{2} \theta {cos}^{2} \theta \bigg)$

We know,

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

$\rm \: = \: 3 \bigg( {sin}^{4}\theta + {cos}^{4}\theta + \cancel{ 2 {sin}^{2}\theta {cos}^{2}\theta} - \cancel{2{sin}^{2} \theta {cos}^{2} \theta} \bigg)$

$\rm \: = \: 3 \bigg( {sin}^{4}\theta + {cos}^{4}\theta \bigg)$

Hence,

$\rm :\longmapsto\:\boxed{ \bf{ \: 2( {sin}^{6}\theta + {cos}^{6}\theta ) + 1 = 3( {sin}^{4}\theta + {cos}^{4}\theta )}}$

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