Question

pls solve it fast its urgent​

Answer 1

${\underline{{\text{To Prove:}}}}$

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

${\underline{{\text{Solution:}}}}$

$\dfrac{ \sin {}^{3} + \cos {}^{3} }{ \sin + \cos } + \dfrac{ \sin {}^{3} - \cos {}^{3} }{ \sin - \cos } \\ \\ [\text{Using} a^3+b^3 = (a+b)(a^2-ab+b^2] \\ \\ [\text{Using} a^3 - b^3 = (a - b)(a^2 + ab+b^2] \\ \\ \dfrac{( \sin + \cos )( { \sin }^{2} - \sin.\cos + \cos {}^{2} ) }{ \sin + \cos } + \dfrac{ (\sin - \cos) ( { \sin }^{2} + \sin.\cos + { \cos}^{2} )}{ \sin - \cos } \\ \\ \dfrac{ \cancel{( \sin + \cos )}( { \sin }^{2} + \cos {}^{2} - \sin.\cos) }{ \cancel{(\sin + \cos)} } + \dfrac{ \cancel{ (\sin - \cos)} ( { \sin }^{2} + { \cos}^{2} + \sin.\cos) }{ \cancel{(\sin - \cos}) } \\ \\ [\text{Using i} ] \\ \\ = 1 - \sin.\cos + 1 + \sin. \cos \\ \\ = 1 + 1 \\ \\ = 2 \\ \\ \therefore \text{L.H.S = R.H.S [Proved}]$

${{\boxed{ \text{\blue{Some Important Trigonometry Identity}}}}}$

$sin \theta . cosec\theta = 1 \\ \\</p><p>\cos \theta. \sec\theta = 1 \\ \\</p><p>\tan\theta . \cot\theta = 1\\ \\</p><p>\sin^2\theta+ \cos^2 \theta= 1 ......[\text{i}] \\ \\</p><p>\cosec^2 \theta -\cot^2 \theta = 1 \\ \\</p><p>\sec^2 \theta - \tan^2 \theta = 1$

Answer 2

To Prove :

$\frac{ {cos}^{3} A + {sin}^{3} A}{cosA+ sinA} + \frac{ {cos}^{3} A - {sin}^{3}A }{cosA - sinA} = 2$

$\huge\red{\mid{\underline{\overline{\textbf {Formula \: used -}}}\mid}}$

• a³+b³ = (a+b) (a²-ab+b²)

• a³-b³ = (a-b) (a²+ab+b²)

• sin²A+cos²A = 1

$\huge{\bold{SoLuTiOn:-}}$

Taking LHS :-

$\frac{ {cos}^{3}A + {sin}^{3}A }{cosA + sinA} + \frac{ {cos}^{3} A - {sin}^{3}A }{cosA- sinA} \\ \\ = \frac{(cosA + sinA)( {cos}^{2}A - cosA.sinA + {sin}^{2}A) }{cosA + sinA} + \frac{(cosA - sinA)( {cos}^{2} A + cosA.sinA + {sin}^{2}A) }{cosA - sinA} \\ \\ = {cos}^{2}A - cosA.sinA + {sin}^{2} A + {cos}^{2} A + cosA.sinA + {sin}^{2} A \\ \\ = ({cos}^{2} A + {sin}^{2} A) + ({cos}^{2} A + {sin}^{2} A) \\ \\ = 1 + 1 \\ \\ = 2 = RHS \\ \\ {\purple{\boxed{\large{\bold{LSH = RHS}}}}}$

$\huge{\bold{ HeNcE \:PrOvEd}}$