Question

In a triangle ABC
Prove that :-
(Cos2A/a²)- (Cos2B/b²) = 1/a² - 1/b²

​

Answer 1

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;

$= > \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = k$

⚫a=k sinA

⚫b=k sinB

⚫c=k sinC

Take L.H.S,

$= > \frac{(cos2A)}{ {a}^{2} } - \frac{(cos2B)}{ {b}^{2} } \\ \\ = > \frac{1 - 2sin {}^{2} a}{ {a}^{2} } - \frac{1 - {sin}^{2}b }{ {b}^{2} } \: \: \: \: [cos2A = 1 - {sin}^{2} A]$

Substituting the values from sine rule into the above equation, we get;

$= > \frac{1 - 2( \frac{a}{k} ) {}^{2} }{ {a}^{2} } - \frac{1 - ( \frac{b}{k}) {}^{2} }{ {b}^{2} } \\ \\ = > \frac{ {k}^{2} - 2 {a}^{2} }{ \frac{ {k}^{2} }{ {a}^{2} } } - \frac{ {k}^{2} - 2 {b}^{2} }{ \frac{ {k}^{2} }{ {b}^{2} } } \\ \\ = > \frac{ {k}^{2} - 2 {a}^{2} }{ {k}^{2} {a}^{2} } - \frac{ {k}^{2} - 2 {b}^{2} }{ {k}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2}( {k}^{2} - 2 {a}^{2}) - {a}^{2} ( {k}^{2} - 2 {b}^{2}) }{ {k}^{2} {a}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2} {k}^{2} - 2 {a}^{2} {b}^{2} - {a}^{2} {k}^{2} + 2 {a}^{2} {b}^{2} }{ {k}^{2} {a}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2} {k}^{2} - {a}^{2} {k}^{2} }{ {k}^{2} {a}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2} - {a}^{2} }{ {a}^{2} {b}^{2} } \\ \\ = > \frac{ {b}^{2} }{ {a}^{2} {b}^{2} } - \frac{ {a}^{2} }{ {a}^{2} {b}^{2} } \\ \\ = > \frac{1}{ {a}^{2} } - \frac{1}{ {b}^{2} } \: \: \: \: \: \: \: \: \: [R.H.S.]$

Hence,

Proved.

Hope it helps ☺️