Question

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Please solve my question :)

If $\frac{sin ^{4} \: A}{a} \: + \: \frac{cos^{4} A}{b} \: = \: \frac{1}{a \: + \: b}$ , then prove that $\frac{sin ^{8} \: A}{ {a}^{3} } \: + \: \frac{cos^{8} \: A}{ {b}^{3} } \: = \: \frac{1}{(a \: + \: b) ^{3} }$​

Answer 1

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$\sf \dfrac{sin^4A}{a}+\dfrac{cos^4A}{b}=\dfrac{1}{a+b}$

• sin²A + cos²A = 1

➠ cos²A = 1 - sin²A

➠ cos⁴A = 1 + sin⁴A - 2sin²A

$\\ \to \sf \dfrac{sin^4A}{a}+\dfrac{1+sin^4A-2sin^2A}{b}=\dfrac{1}{a+b}$

$\\\to \sf bsin^4A+a+asin^4A-2asin^2A=\dfrac{ab}{a+b}$

$\to \sf sin^4A(a+b)-2asin^2A+a=\dfrac{ab}{a+b}$

$\\ \to \sf sin^4A(a+b)^2-2(sin^2A)(a+b)a+a(a+b)=ab$

$\\\to \sf sin^4A(a+b)^2-2[sin^2A(a+b)]a+a^2+ab-ab=0$

$\\\to \sf [sin^2A(a+b)]^2-2[sin^2A(a+b)]a+a^2=0$

• (a - b)² = a² - 2ab + b²

$\\\to \sf [sin^2A(a+b)-a]^2=0$

$\\\to \blue{\textsf{\textbf{ sin ^\text{2}\text{A}=\dfrac{\text a}{\text{a\ +\ b}} }}}$

Now , cos²A = 1 - sin²A

$\\\to \sf cos^2A=1-\dfrac{a}{a+b}$

$\\\to \green{\textsf{\textbf{ \text{cos}^\text{2}\text{A}=\dfrac{\text{b}}{\text{a\ +\ b}} }}}$

So ,

$\\\sf \dfrac{\text{sin}^\text{8}A}{a^3}+\dfrac{cos^8A}{b^3}$

$\\\to \sf \dfrac{(sin^2A)^4}{a^3}+\dfrac{(cos^2A)^4}{b^3}$

$\\\to \sf \dfrac{\big(\frac{a}{a+b}\big)^{\tiny{4}}}{a^3}+\dfrac{\big(\frac{b}{a+b}\big)^{\tiny{4}}}{b^3}$

$\\\to \sf \dfrac{a}{(a+b)^4}+\dfrac{b}{(a+b)^4}$

$\\\to \sf \dfrac{a+b}{(a+b)^4}$

$\\\pink{\to \textsf{\textbf{ \dfrac{\text 1}{\text{(a\ +\ b) ^\text{3} }} }}}$

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Answer 2

$\sf \dfrac{sin^4A}{a}+\dfrac{cos^4A}{b}=\dfrac{1}{a+b}$

$\begin{gathered}\\ \to \sf \dfrac{sin^4A}{a}+\dfrac{1+sin^4A-2sin^2A}{b}=\dfrac{1}{a+b}\\\end{gathered}$

$\begin{gathered}\\\to \sf bsin^4A+a+asin^4A-2asin^2A=\dfrac{ab}{a+b}\end{gathered}$

$\begin{gathered}\to \sf sin^4A(a+b)-2asin^2A+a=\dfrac{ab}{a+b}\\\end{gathered}$

$\begin{gathered}\\ \to \sf sin^4A(a+b)^2-2(sin^2A)(a+b)a+a(a+b)=ab\\\end{gathered}$

$\begin{gathered}\\\to \sf sin^4A(a+b)^2-2[sin^2A(a+b)]a+a^2+ab-ab=0\\\end{gathered}$

$\begin{gathered}\\\to \sf [sin^2A(a+b)]^2-2[sin^2A(a+b)]a+a^2=0\\\end{gathered}$

$\begin{gathered}\\\to \sf [sin^2A(a+b)-a]^2=0\\\end{gathered}$

$\begin{gathered}\\\to {\textsf{\textbf{ sin ^\text{2}\text{A}=\dfrac{\text a}{\text{a\ +\ b}} }}}\\\end{gathered}$

Now , cos²A = 1 - sin²A

$\begin{gathered}\\\to \sf cos^2A=1-\dfrac{a}{a+b}\\\end{gathered}$

$\begin{gathered}\\\to {\textsf{\textbf{ \text{cos}^\text{2}\text{A}=\dfrac{\text{b}}{\text{a\ +\ b}} }}}\\\end{gathered}$

$\begin{gathered}\\\sf \dfrac{\text{sin}^\text{8}A}{a^3}+\dfrac{cos^8A}{b^3}\\\end{gathered}$

$\begin{gathered}\\\to \sf \dfrac{(sin^2A)^4}{a^3}+\dfrac{(cos^2A)^4}{b^3}\\\end{gathered}$

$\begin{gathered}\\\to \sf \dfrac{\big(\frac{a}{a+b}\big)^{\tiny{4}}}{a^3}+\dfrac{\big(\frac{b}{a+b}\big)^{\tiny{4}}}{b^3}\\\end{gathered}$

$\begin{gathered}\\\to \sf \dfrac{a}{(a+b)^4}+\dfrac{b}{(a+b)^4}\\\end{gathered}$

$\begin{gathered}\\\to \sf \dfrac{a+b}{(a+b)^4}\\\end{gathered}$

$\begin{gathered}{\to \textsf{\textbf{ \dfrac{\text 1}{\text{(a\ +\ b) ^\text{3} }} }}}\\\end{gathered}$