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B-y-expessing-each-side-of-the-equation-in-terms-of-tanA-or-otherwise-show-that-sin2A-cos2A-1-sin2A-cos2A-1-tan-45-A-tanA-

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Question Number 8297 by lepan last updated on 06/Oct/16
B_ y expessing each side of the equation in terms of tanA ,or otherwise show that ((sin2A+cos2A+1)/(sin2A+cos2A−1))=((tan(45°+A))/(tanA))
$$\underset{} {{B}y}\:{expessing}\:{each}\:{side}\:{of}\:{the} \\ $$$${equation}\:{in}\:{terms}\:{of}\:{tanA}\:,{or}\: \\ $$$${otherwise}\:{show}\:{that} \\ $$$$\frac{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}+\mathrm{1}}{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}−\mathrm{1}}=\frac{{tan}\left(\mathrm{45}°+{A}\right)}{{tanA}} \\ $$
Answered by Rasheed Soomro last updated on 07/Oct/16
((sin2A+cos2A+1)/(sin2A+cos2A−1))=((tan(45°+A))/(tanA)) LHS_(−) : ((sin2A+cos2A+1)/(sin2A+cos2A−1)) sin2A=((2tanA)/(1+tan^2 A)) cos2A=((1−tan^2 A)/(1+tan^2 A)) =((((2tanA)/(1+tan^2 A))+((1−tan^2 A)/(1+tan^2 A))+1)/(((2tanA)/(1+tan^2 A))+((1−tan^2 A)/(1+tan^2 A))−1)) =(((2tanA+1−tan^2 A+1+tan^2 A)/(1+tan^2 A))/((2tanA+1−tan^2 A−1−tan^2 A)/(1+tan^2 A))) =((2tanA+1−tan^2 A+1+tan^2 A)/(2tanA+1−tan^2 A−1−tan^2 A)) =((2tanA+2)/(2tanA−2tan^2 A)) =((tanA+1)/(tanA−tan^2 A)) =((1+tanA)/(tanA(1−tanA))) RHS_(−) : ((tan(45°+A))/(tanA)) =(((tan45+tanA)/(1−tan45 tanA))/(tanA)) =(((1+tanA)/(1− tanA))/(tanA)) =((1+tanA)/(tanA(1−tanA))) LHS=RHS Proved
$$\frac{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}+\mathrm{1}}{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}−\mathrm{1}}=\frac{{tan}\left(\mathrm{45}°+{A}\right)}{{tanA}} \\ $$$$ \\ $$$$\underset{−} {\boldsymbol{\mathrm{LHS}}}:\:\:\frac{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}+\mathrm{1}}{{sin}\mathrm{2}{A}+{cos}\mathrm{2}{A}−\mathrm{1}} \\ $$$${sin}\mathrm{2}{A}=\frac{\mathrm{2}{tanA}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}\:\:\: \\ $$$${cos}\mathrm{2}{A}=\frac{\mathrm{1}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}} \\ $$$$=\frac{\frac{\mathrm{2}{tanA}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}+\mathrm{1}}{\frac{\mathrm{2}{tanA}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}−\mathrm{1}} \\ $$$$=\frac{\frac{\mathrm{2}{tanA}+\mathrm{1}−{tan}^{\mathrm{2}} {A}+\mathrm{1}+{tan}^{\mathrm{2}} {A}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}}{\frac{\mathrm{2}{tanA}+\mathrm{1}−{tan}^{\mathrm{2}} {A}−\mathrm{1}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}} \\ $$$$=\frac{\mathrm{2}{tanA}+\mathrm{1}−{tan}^{\mathrm{2}} {A}+\mathrm{1}+{tan}^{\mathrm{2}} {A}}{\mathrm{2}{tanA}+\mathrm{1}−{tan}^{\mathrm{2}} {A}−\mathrm{1}−{tan}^{\mathrm{2}} {A}} \\ $$$$=\frac{\mathrm{2}{tanA}+\mathrm{2}}{\mathrm{2}{tanA}−\mathrm{2}{tan}^{\mathrm{2}} {A}} \\ $$$$=\frac{{tanA}+\mathrm{1}}{{tanA}−{tan}^{\mathrm{2}} {A}} \\ $$$$=\frac{\mathrm{1}+{tanA}}{{tanA}\left(\mathrm{1}−{tanA}\right)} \\ $$$$ \\ $$$$\underset{−} {\boldsymbol{\mathrm{RHS}}}:\:\:\frac{{tan}\left(\mathrm{45}°+{A}\right)}{{tanA}} \\ $$$$=\frac{\frac{{tan}\mathrm{45}+{tanA}}{\mathrm{1}−{tan}\mathrm{45}\:{tanA}}}{{tanA}} \\ $$$$=\frac{\frac{\mathrm{1}+{tanA}}{\mathrm{1}−\:{tanA}}}{{tanA}} \\ $$$$=\frac{\mathrm{1}+{tanA}}{{tanA}\left(\mathrm{1}−{tanA}\right)} \\ $$$$\mathrm{LHS}=\mathrm{RHS} \\ $$$$\mathrm{Proved} \\ $$

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