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Question Number 73160 by Aditya789 last updated on 07/Nov/19

1+tanAtan(A/2)=tanAcot(A/2)−1=secA

$$\mathrm{1}+{tanAtan}\frac{{A}}{\mathrm{2}}={tanAcot}\frac{{A}}{\mathrm{2}}−\mathrm{1}={secA} \\ $$

Answered by $@ty@m123 last updated on 07/Nov/19

1+tan Atan (A/2) =1+((2tan (A/2))/(1−tan^2 (A/2))) .tan (A/2) =1+((2tan^2 (A/2))/(1−tan^2 (A/2))) =((1−tan^2 (A/2)+2tan^2 (A/2))/(1−tan^2 (A/2))) =((1+tan^2 (A/2))/(1−tan^2 (A/2))) =(1/(((1−tan^2 (A/2))/(1+tan^2 (A/2))) )) =(1/(cos A)) =sec A Second Part: Try yourself in similar manner.

$$\mathrm{1}+\mathrm{tan}\:{A}\mathrm{tan}\:\frac{{A}}{\mathrm{2}} \\ $$$$=\mathrm{1}+\frac{\mathrm{2tan}\:\frac{{A}}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}\:.\mathrm{tan}\:\frac{{A}}{\mathrm{2}} \\ $$$$=\mathrm{1}+\frac{\mathrm{2tan}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}\: \\ $$$$=\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}+\mathrm{2tan}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}\: \\ $$$$=\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}\: \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}}\:} \\ $$$$=\frac{\mathrm{1}}{\mathrm{cos}\:{A}} \\ $$$$=\mathrm{sec}\:{A} \\ $$$${Second}\:{Part}:\:{Try}\:{yourself}\:{in}\:{similar} \\ $$$${manner}. \\ $$

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