Question

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

Answer 1

`Tan^2x= (C -A)/((B-C))`

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`

⇒
`(Acos^2x)/(cos^2x) + (Bsin^2x)/(cos^2x) = (C)/(cos^2x)`

⇒
`A + B((sin^2x)/(cos^2x)) = C(1)/(cos^2x)` {
`(sin^2x)/(cos^2x) = Tan^2x , (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= (C -A)/((B-C))`

Therefore ,
`Tan^2x= (C -A)/((B-C))` .