Question

Acos²theta + Bsin²theta = C
Show that
tan²theta =C- A/B-C​

Answer 1

$Tan^2x= \frac{C -A}{(B-C)}$

Step-by-step explanation:

Here we have  , Acos²theta + Bsin²theta = C or , $Acos^2theta + Bsin^2theta = C$

Let theta = x , So  $Acos^2x + Bsin^2x = C$ . Let's solve it further

⇒  $Acos^2x + Bsin^2x = C$

⇒  $\frac{Acos^2x}{cos^2x} + \frac{Bsin^2x}{cos^2x} = \frac{C}{cos^2x}$

⇒  $A + B(\frac{sin^2x}{cos^2x}) = C\frac{1}{cos^2x}$            { $\frac{sin^2x}{cos^2x} = Tan^2x , \frac{1}{cos^2x} = sec^2x$  }

⇒  $A + B(Tan^2x) = C(sec^2x)$     {  $sec^2x=1+Tan^2x$  }

⇒  $A + B(Tan^2x) = C( 1+Tan^2x )$

⇒  $A + B(Tan^2x) = C+C(Tan^2x )$

⇒  $B(Tan^2x)-C(Tan^2x ) = C -A$

⇒  $(B-C)(Tan^2x)= C -A$

⇒  $Tan^2x= \frac{C -A}{(B-C)}$

Therefore ,  $Tan^2x= \frac{C -A}{(B-C)}$ .