Question

this is official question

dont give me wrong answer if u dont know

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Answer 1

${\huge{\rm{\underline{\underline{Answer\checkmark}}}}}$

$\rm \: \star \: refer \: to \: the \: attachment$

$\large \: \rm \: formula \: used :$

$\mapsto \rm \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + {y}^{2} + xy)$

$\mapsto \rm \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$

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

Answer 2

2

Step-by-step explanation:

$\begin{lgathered}Concept:\\a^3+b^3=(a+b)(a^2-ab+b^2)\\a^3-b^3=(a-b)(a^2+ab+b^2)\end{lgathered} </p><p>Concept:$

$\begin{lgathered}\frac{cos^3\theta+sin^3\theta}{cos\theta+sin\theta}+\frac{cos^3\theta-sin^3\theta}{cos\theta-sin\theta}\\\\=\frac{(cos\theta+sin\theta)(cos^2\theta-cos\theta.sin\theta+sin^2\theta)}{cos\theta+sin\theta}+\frac{(cos\theta-sin\theta)(cos^2\theta+cos\theta.sin\theta+sin^2\theta)}{cos\theta-sin\theta}\\\\=(cos^2\theta-cos\theta.sin\theta+sin^2\theta)+(cos^2\theta+cos\theta.sin\theta+sin^2\theta)\\\\=(cos^2\theta+sin^2\theta)+(cos^2\theta+sin^2\theta)\\\\=1+1= 2\end{lgathered}$