solve 14th one,with proper explanation.
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Answers
Step-by-step explanation:
\begin{gathered}ans \\ \\ given \: \: \: \frac{ {sin}^{4} \alpha }{a} + \frac{ {cos}^{4} \alpha }{b} = \frac{1}{a + b} \\ \\ or \: \: \frac{ { ({sin}^{2} \alpha ) }^{2} }{a} + \frac{ {(1 - {sin}^{2} \alpha )}^{2} }{b} \\ \\ let \: us \: consider \: {sin}^{2} \alpha = x \\ \\ then \: we \: get \\ \\ \frac{ {x}^{2} }{a} + \frac{ {(1 - x)}^{2} }{b} = \frac{1}{a + b} \\ \\ or \: \: (a + b)(b {x}^{2} + a(1 - 2x + {x}^{2} )) = ab \\ \\ or \: \: {(a + b)}^{2} {x}^{2} - 2a(a + b)x + {a}^{2} + ab = ab \\ \\ or \: \: {(a + b)}^{2} {x}^{2} - 2a( a+ b)x + {a}^{2} = 0 \\ \\ or \: \:{ ((a + b)x - a)}^{2} = 0 \\ \\ thus \: \: (a + b)x - a = 0 \\ \\ or \: \: x = \frac{a}{a + b} \\ \\ so \: \: {sin}^{2} \alpha = \frac{a}{a + b} \\ \\ hence \: \: {cos}^{2} \alpha = 1 - {sin}^{2} \alpha \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1 - {( \frac{a}{a + b}) } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{a + b - a}{a + b} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{b}{a + b} \\ \\ now \: \: \: lhs \\ \\ = \frac{ {sin}^{8} \alpha }{ {a}^{3} } + \frac{ {cos}^{8} \alpha }{ {b}^{3} } \\ \\ = \frac{1}{ {a}^{3} } {( {sin}^{2} \alpha ) }^{4} + \frac{1}{ {b}^{3} } { {(cos}^{2} \alpha ) }^{4} \\ \\ = \frac{1}{ {a}^{3} } \frac{ {a}^{4} }{( {a + b})^{4} } + \frac{1}{ {b}^{3} } \frac{ {b}^{4} }{ {(a + b)}^{4} } \\ \\ = \frac{a}{ {(a + b)}^{4} } + \frac{b}{ {(a + b)}^{4} } \\ \\ = \frac{a + b}{ {(a + b)}^{4} } \\ \\ = \frac{1}{ {(a + b)}^{3} } \\ \\ = rhs \: \: (proved)\end{gathered}
ans
given
a
sin
4
α
+
b
cos
4
α
=
a+b
1
or
a
(sin
2
α)
2
+
b
(1−sin
2
α)
2
letusconsidersin
2
α=x
thenweget
a
x
2
+
b
(1−x)
2
=
a+b
1
or(a+b)(bx
2
+a(1−2x+x
2
))=ab
or(a+b)
2
x
2
−2a(a+b)x+a
2
+ab=ab
or(a+b)
2
x
2
−2a(a+b)x+a
2
=0
or((a+b)x−a)
2
=0
thus(a+b)x−a=0
orx=
a+b
a
sosin
2
α=
a+b
a
hencecos
2
α=1−sin
2
α
=1−(
a+b
a
)
=
a+b
a+b−a
=
a+b
b
nowlhs
=
a
3
sin
8
α
+
b
3
cos
8
α
=
a
3
1
(sin
2
α)
4
+
b
3
1
(cos
2
α)
4
=
a
3
1
(a+b)
4
a
4
+
b
3
1
(a+b)
4
b
4
=
(a+b)
4
a
+
(a+b)
4
b
=
(a+b)
4
a+b
=
(a+b)
3
1
=rhs(proved)
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