Math, asked by nishu365592, 9 months ago

Prove that
Cos³theta + sin³theta / cos theta + sin theta + cos³theta-sin³theta/ cos theta - sin theta =2​

Attachments:

Answers

Answered by Anonymous
14

Step-by-step explanation:

To Prove :

 \dfrac{ {cos}^{3}\theta  +   { \sin}^{3} \theta }{ \cos\theta  +  \sin \theta}  +  \dfrac{ {cos}^{3}\theta   -   { \sin}^{3} \theta }{ \cos\theta   -   \sin \theta} = 2

L.H.S. =

 \dfrac{ {cos}^{3}\theta  +   { \sin}^{3} \theta }{ \cos\theta  +  \sin \theta}  +  \dfrac{ {cos}^{3}\theta   -   { \sin}^{3} \theta }{ \cos\theta   -   \sin \theta}

 \dfrac{( {cos}\theta )^{3}   +   ({ \sin} \theta)^{3}  }{ \cos\theta  +  \sin \theta}  +  \dfrac{( {cos}\theta)^{3}    -   { (\sin} \theta ) ^{3} }{ \cos\theta   -   \sin \theta}

{\boxed{\red{Identity \ : \ a^3 + b^3 = (a + b)(a^2 + b^2 - ab)}}}

{\boxed{\red{Identity \ : \ a^3 - b^3 = (a - b)(a^2 + b^2 + ab)}}}

 \dfrac{(cos\theta + sin\theta)( {cos}^{2} \theta +  {sin}^{2} \theta - cos\theta \: sin\theta)}{cos\theta + sin\theta \: }  +  \dfrac{(cos\theta  -  sin\theta)( {cos}^{2} \theta +  {sin}^{2} \theta  +  cos\theta \: sin\theta)}{cos\theta  -  sin\theta \: }

 \dfrac{\cancel{(cos\theta + sin\theta)}( {cos}^{2} \theta +  {sin}^{2} \theta - cos\theta \: sin\theta)}{\cancel{cos\theta + sin\theta \: }}  +  \dfrac{\cancel{(cos\theta  -  sin\theta)}( {cos}^{2} \theta +  {sin}^{2} \theta  +  cos\theta \: sin\theta)}{\cancel{cos\theta  -  sin\theta \: }}

→ cos²θ + sin²θ - cosθsinθ + cos²θ + sin²θ + cosθsinθ

→ cos²θ + sin²θ + cos²θ + sin²θ

→ 2cos²θ + 2sin²θ

→ 2(cos²θ + sin²θ)

{\boxed{\red{cos^2 \theta + sin^2 \theta = 1}}}

→ 2(1)

→ 2

= R.H.S.

Hence, proved !!

Answered by nandanasnair2005
6

Answer:

Answer in the attachment

Attachments:
Similar questions