In A ABC Prove that
COS2A
cos2B
1
1
1
a?
02
a ?
b2
Answers
Answer:
SOLUTION:-
Given:
In a ∆ABC,
To prove:
(cos2A/a²)- (cos2B/b²)=1/a² - 1/b²
Proof:
Let a,b & c be the sides of any ∆ABC.
Then by applying the sine rule, we get;
= > \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = k=>
sinA
a
=
sinB
b
=
sinC
c
=k
⚫a=k sinA
⚫b=k sinB
⚫c=k sinC
Take L.H.S,
\begin{gathered} = > \frac{(cos2A)}{ {a}^{2} } - \frac{(cos2B)}{ {b}^{2} } \\ \\ = > \frac{1 - 2sin {}^{2} a}{ {a}^{2} } - \frac{1 - {sin}^{2}b }{ {b}^{2} } \: \: \: \: [cos2A = 1 - {sin}^{2} A]\end{gathered}
=>
a
2
(cos2A)
−
b
2
(cos2B)
=>
a
2
1−2sin
2
a
−
b
2
1−sin
2
b
[cos2A=1−sin
2
A]
Substituting the values from sine rule into the above equation, we get;
\begin{gathered} = > \frac{1 - 2( \frac{a}{k} ) {}^{2} }{ {a}^{2} } - \frac{1 - ( \frac{b}{k}) {}^{2} }{ {b}^{2} } \\ \\ = > \frac{ {k}^{2} - 2 {a}^{2} }{ \frac{ {k}^{2} }{ {a}^{2} } } - \frac{ {k}^{2} - 2 {b}^{2} }{ \frac{ {k}^{2} }{ {b}^{2} } } \\ \\ = > \frac{ {k}^{2} - 2 {a}^{2} }{ {k}^{2} {a}^{2} } - \frac{ {k}^{2} - 2 {b}^{2} }{ {k}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2}( {k}^{2} - 2 {a}^{2}) - {a}^{2} ( {k}^{2} - 2 {b}^{2}) }{ {k}^{2} {a}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2} {k}^{2} - 2 {a}^{2} {b}^{2} - {a}^{2} {k}^{2} + 2 {a}^{2} {b}^{2} }{ {k}^{2} {a}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2} {k}^{2} - {a}^{2} {k}^{2} }{ {k}^{2} {a}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2} - {a}^{2} }{ {a}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2} }{ {a}^{2} {b}^{2} } - \frac{ {a}^{2} }{ {a}^{2} {b}^{2} } \\ \\ = > \frac{1}{ {a}^{2} } - \frac{1}{ {b}^{2} } \: \: \: \: \: \: \: \: \: [R.H.S.]\end{gathered}
=>
a
2
1−2(
k
a
)
2
−
b
2
1−(
k
b
)
2
=>
a
2
k
2
k
2
−2a
2
−
b
2
k
2
k
2
−2b
2
=>
k
2
a
2
k
2
−2a
2
−
k
2
b
2
k
2
−2b
2
=>
k
2
a
2
b
2
b
2
(k
2
−2a
2
)−a
2
(k
2
−2b
2
)
=>
k
2
a
2
b
2
b
2
k
2
−2a
2
b
2
−a
2
k
2
+2a
2
b
2
=>
k
2
a
2
b
2
b
2
k
2
−a
2
k
2
=>
a
2
b
2
b
2
−a
2
=>
a
2
b
2
b
2
−
a
2
b
2
a
2
=>
a
2
1
−
b
2
1
[R.H.S.]
Hence,
Proved.
Hope it helps ☺️