[tex]\Large {\underline{\sf{Solution :-}}}[/tex]
[tex]{\underline{\sf{To \: prove :}}} \: \: \sf cos \: 2\phi = \dfrac{cos \: 2\theta + e}{1 + e \: cos \: 2\theta}[/tex]
[tex]\sf Given : \: tan \: \phi = \sqrt{\dfrac{1 - e}{1 + e}} \: tan \: \theta[/tex]
[tex]\longrightarrow \: {\underline{\boxed{\sf{cos \: 2\phi = \dfrac{1 - {tan}^{2} \phi}{1 + {tan}^{2} \phi}}}}}[/tex]
[tex]\sf cos \: 2\phi = \dfrac{1 - \dfrac{1 - e}{1 + e} \: \dfrac{{sin}^{2} \theta}{{cos}^{2} \theta}}{1 + \dfrac{1 - e}{1 + e} \: \dfrac{{sin}^{2} \theta}{{cos}^{2} \theta}}[/tex]
[tex]\sf cos \: 2\phi= \dfrac{{cos}^{2} \theta + e \: {cos}^{2} \theta - (1 - e) {sin}^{2} \theta}{{cos}^{2} \theta + e \: {cos}^{2} \theta + (1 - e) {sin}^{2} \theta}[/tex]
[tex]\sf cos \: 2\phi = \dfrac{({cos}^{2} \theta - {sin}^{2} \theta) + e({sin}^{2} \theta + {cos}^{2} \theta)}{({sin}^{2} \theta + {cos}^{2} \theta) + e({cos}^{2} \theta - {sin}^{2} \theta)}[/tex]
[tex]\longmapsto \: {\underline{\overline{\boxed{\sf{cos \: 2\phi = \dfrac{cos \: 2\theta + e}{1 + e \: cos \: 2\theta}}}}}}[/tex]
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[tex]\Large {\underline{\sf{Solution :-}}}[/tex]
[tex]{\underline{\sf{To \: prove :}}} \: \: \sf cos \: 2\phi = \dfrac{cos \: 2\theta + e}{1 + e \: cos \: 2\theta}[/tex]
[tex]\sf Given : \: tan \: \phi = \sqrt{\dfrac{1 - e}{1 + e}} \: tan \: \theta[/tex]
[tex]\longrightarrow \: {\underline{\boxed{\sf{cos \: 2\phi = \dfrac{1 - {tan}^{2} \phi}{1 + {tan}^{2} \phi}}}}}[/tex]
[tex]\sf cos \: 2\phi = \dfrac{1 - \dfrac{1 - e}{1 + e} \: \dfrac{{sin}^{2} \theta}{{cos}^{2} \theta}}{1 + \dfrac{1 - e}{1 + e} \: \dfrac{{sin}^{2} \theta}{{cos}^{2} \theta}}[/tex]
[tex]\sf cos \: 2\phi= \dfrac{{cos}^{2} \theta + e \: {cos}^{2} \theta - (1 - e) {sin}^{2} \theta}{{cos}^{2} \theta + e \: {cos}^{2} \theta + (1 - e) {sin}^{2} \theta}[/tex]
[tex]\sf cos \: 2\phi = \dfrac{({cos}^{2} \theta - {sin}^{2} \theta) + e({sin}^{2} \theta + {cos}^{2} \theta)}{({sin}^{2} \theta + {cos}^{2} \theta) + e({cos}^{2} \theta - {sin}^{2} \theta)}[/tex]
[tex]\longmapsto \: {\underline{\overline{\boxed{\sf{cos \: 2\phi = \dfrac{cos \: 2\theta + e}{1 + e \: cos \: 2\theta}}}}}}[/tex]