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Question Number 149 by rajabhay last updated on 25/Jan/15
if tan A=(√2)−1 then find ((tan A)/(1+tan^2 A)) = ?
$$\mathrm{if}\:\mathrm{tan}\:\mathrm{A}=\sqrt{\mathrm{2}}−\mathrm{1}\:\mathrm{then}\:\mathrm{find} \\ $$$$\frac{\mathrm{tan}\:\mathrm{A}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{A}}\:=\:? \\ $$
Answered by prakash jain last updated on 12/Dec/14
tan^2 A=((√2)−1)^2 =3−2(√2) ((tan A)/(1+tan^2 A))=(((√2)−1)/(4−2(√2)))=((((√2)−1)(4+2(√2)))/((4−2(√2))(4+2(√2)))) =((4(√2)−4+4−2(√2))/(16−8))=((2(√2))/8)=((√2)/4)
$$\mathrm{tan}^{\mathrm{2}} \mathrm{A}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{tan}\:{A}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {A}}=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\right)}{\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{4}+\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{16}−\mathrm{8}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

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