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Question-27850




Question Number 27850 by v7277668420 last updated on 15/Jan/18
Answered by bmind4860 last updated on 15/Jan/18
a^2 sec^2 θ−b^2 (sec^2 θ−1)=c^2   after rearranging we have  sec^2 θ=((c^2 −b^2 )/(a^2 −b^2 ))  ⇒1−sin^2 θ=((a^2 −b^2 )/(c^2 −b^2 ))  ⇒sin^2 θ=((c^2 −a^2 )/(c^2 −b^2 ))  sinθ=±(√((c^2 −a^2 )/(c^2 −b^2 )))
$${a}^{\mathrm{2}} {sec}^{\mathrm{2}} \theta−{b}^{\mathrm{2}} \left({sec}^{\mathrm{2}} \theta−\mathrm{1}\right)={c}^{\mathrm{2}} \\ $$$${after}\:{rearranging}\:{we}\:{have} \\ $$$${sec}^{\mathrm{2}} \theta=\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{1}−{sin}^{\mathrm{2}} \theta=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{sin}^{\mathrm{2}} \theta=\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${sin}\theta=\pm\sqrt{\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }} \\ $$

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