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stepan [7]
3 years ago
9

Acos²theta + Bsin²theta = CShow thattan²theta =C- A/B-C​

Mathematics
1 answer:
olchik [2.2K]3 years ago
4 0

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

<u>Step-by-step explanation:</u>

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)} .

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I think your question is missed of key information, allow me to add in and hope it will fit the original one.  

Please have a look at the attached photo.  

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