1.

Solve this..............​

Answer»

♣ɢɪᴠᴇɴ:–

\large \bf \dfrac {sin ^{4} \: A}{a} \: + \: \frac{cos^{4} A}{b} \: = \: \frac{<klux>1</klux>}{a \: + \: b} \\

♣ᴛᴏ ᴘʀᴏᴠᴇ:–

\large \bf \dfrac{sin ^{8} \: A}{ {a}^{3} } \: + \: \frac{cos^{8} \: A}{ {b}^{3} } \: = \: \frac{1}{(a \: + \: b) ^{3} } \\

♣sᴏʟᴜᴛɪᴏɴ:–

➤ cos²A = 1 - sin²A

➤ cos⁴A = 1 + sin⁴A - 2sin²A

 cos⁴A = 1 + sin⁴A - 2sin²A

\begin{gathered}\\ \to \sf \dfrac{sin^4A}{a}+\dfrac{1+sin^4A-2sin^2A}{b}=\dfrac{1}{a+b}\\\end{gathered}\\\begin{gathered}\\\to \sf bsin^4A+a+asin^4A-2asin^2A=\dfrac{ab}{a+b}\end{gathered}\\\begin{gathered}\to \sf sin^4A(a+b)-2asin^2A+a=\dfrac{ab}{a+b}\\\end{gathered}\\\begin{gathered}\\ \to \sf sin^4A(a+b)^2-2(sin^2A)(a+b)a+a(a+b)=ab\\\end{gathered}\\\begin{gathered}\\\to \sf sin^4A(a+b)^2-2[sin^2A(a+b)]a+a^2+ab-ab=0\\\end{gathered}\\\begin{gathered}\\\to \sf[sin^2A(a+b)]^2-2[sin^2A(a+b)]a+a^2=0\\\end{gathered}\\(a - b)² = a² - 2ab + b²\\\begin{gathered}\\\to \sf [sin^2A(a+b)-a]^2=0\\\end{gathered}\\\begin{gathered}\\\to \purple{\textsf{\textbf{ sin$^\text{2}\text{A}=\dfrac{\text a}{\text{a\ +\ b}} $}}}\\\end{gathered}\\\bull \: \bf{Now , cos²A = 1 - sin²A}\\\begin{gathered}\\\to \sf cos^2A=1-\dfrac{a}{a+b}\\\end{gathered}\\\begin{gathered}\\\to \pink{\textsf{\textbf{$\text{cos}^\text{2}\text{A}=\dfrac{\text{b}}{\text{a\ +\ b}}$}}}\\\end{gathered}\\\bull \: \bf{So} ,\\\begin{gathered}\\\sf \dfrac{\text{sin}^\text{8}A}{a^3}+\dfrac{cos^8A}{b^3}\\\end{gathered}\\\begin{gathered}\\\to \sf \dfrac{(sin^2A)^4}{a^3}+\dfrac{(cos^2A)^4}{b^3}\\\end{gathered}\\\begin{gathered}\\\to \sf \dfrac{\big(\dfrac{a}{a+b}\big)^{\tiny{4}}}{a^3}+\dfrac{\big(\dfrac{b}{a+b}\big)^{\tiny{4}}}{b^3}\\\end{gathered}\\\begin{gathered}\\\to\sf \dfrac{a}{(a+b)^4}+\dfrac{b}{(a+b)^4}\\\end{gathered}\\\begin{gathered}\\\to \sf \dfrac{a+b}{(a+b)^4}\\\end{gathered}\\\begin{gathered}\\\red{\leadsto \textsf{\textbf{$ \dfrac{\text 1}{\text{(a\ +\ b)$^\text{3}$}} $}}}\\\end{gathered} \\


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