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Question Number 186590 by mustafazaheen last updated on 06/Feb/23

A+B=(π/4) find the (1+tanA)(1+tanB)=? with explanotory solution

$$\mathrm{A}+\mathrm{B}=\frac{\pi}{\mathrm{4}}\:\:\:\:\:{find}\:{the}\:\left(\mathrm{1}+\mathrm{tan}{A}\right)\left(\mathrm{1}+\mathrm{tan}{B}\right)=?\: \\ $$$${with}\:{explanotory}\:{solution} \\ $$

Answered by mr W last updated on 06/Feb/23

tan (A+B)=tan (π/4)=1 ((tan A+tan B)/(1−tan A tan B))=1 tan A+tan B=1−tan A tan B 1+tan A+tan B+tan A tan B=2 1+tan A+(1+tan A)tan B=2 (1+tan A)(1+tan B)=2 ✓

$$\mathrm{tan}\:\left({A}+{B}\right)=\mathrm{tan}\:\frac{\pi}{\mathrm{4}}=\mathrm{1} \\ $$$$\frac{\mathrm{tan}\:{A}+\mathrm{tan}\:{B}}{\mathrm{1}−\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}}=\mathrm{1} \\ $$$$\mathrm{tan}\:{A}+\mathrm{tan}\:{B}=\mathrm{1}−\mathrm{tan}\:{A}\:\mathrm{tan}\:{B} \\ $$$$\mathrm{1}+\mathrm{tan}\:{A}+\mathrm{tan}\:{B}+\mathrm{tan}\:{A}\:\mathrm{tan}\:{B}=\mathrm{2} \\ $$$$\mathrm{1}+\mathrm{tan}\:{A}+\left(\mathrm{1}+\mathrm{tan}\:{A}\right)\mathrm{tan}\:{B}=\mathrm{2} \\ $$$$\left(\mathrm{1}+\mathrm{tan}\:{A}\right)\left(\mathrm{1}+\mathrm{tan}\:{B}\right)=\mathrm{2}\:\checkmark \\ $$

Answered by Frix last updated on 06/Feb/23

1+tan B =1+((sin B)/(cos B))=1+((sin ((π/4)−A))/(cos ((π/4)−A)))= =1+(((cos A −sin A)/( (√2)))/((cos A +sin A)/( (√2))))=1+((cos A −sin A)/(cos A +sin A))= =((2cos x)/(cos x +sin x))=(2/(1+tan A)) ⇒ answer is2

$$\mathrm{1}+\mathrm{tan}\:{B}\:=\mathrm{1}+\frac{\mathrm{sin}\:{B}}{\mathrm{cos}\:{B}}=\mathrm{1}+\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{A}\right)}{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−{A}\right)}= \\ $$$$=\mathrm{1}+\frac{\frac{\mathrm{cos}\:{A}\:−\mathrm{sin}\:{A}}{\:\sqrt{\mathrm{2}}}}{\frac{\mathrm{cos}\:{A}\:+\mathrm{sin}\:{A}}{\:\sqrt{\mathrm{2}}}}=\mathrm{1}+\frac{\mathrm{cos}\:{A}\:−\mathrm{sin}\:{A}}{\mathrm{cos}\:{A}\:+\mathrm{sin}\:{A}}= \\ $$$$=\frac{\mathrm{2cos}\:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}\:{A}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is2} \\ $$

Answered by a.lgnaoui last updated on 06/Feb/23

(1+tanA)(1+tanB)=P P=1+(tanA+tanB )+tan A×tanB we know that tan (A+B)=((tan A+tan B)/(1−tan A×tanB)) tanA+tanB=tan (A+B)[1−tan A×tan B] =1−tan A×tan B P=1+1−tanA×tanB+tan A×tan B =2 donc (1+tan A)(1+tan B)=2

$$\left(\mathrm{1}+\mathrm{tan}{A}\right)\left(\mathrm{1}+\mathrm{tan}{B}\right)=\boldsymbol{{P}} \\ $$$$\boldsymbol{{P}}=\mathrm{1}+\left(\mathrm{tan}{A}+\mathrm{tan}{B}\:\right)+\mathrm{tan}\:{A}×\mathrm{tan}{B} \\ $$$${we}\:{know}\:{that}\:\:\:\mathrm{tan}\:\left({A}+{B}\right)=\frac{\mathrm{tan}\:{A}+\mathrm{tan}\:{B}}{\mathrm{1}−\mathrm{tan}\:{A}×\mathrm{tan}{B}} \\ $$$$\mathrm{tan}{A}+\mathrm{tan}{B}=\mathrm{tan}\:\left({A}+{B}\right)\left[\mathrm{1}−\mathrm{tan}\:{A}×\mathrm{tan}\:{B}\right] \\ $$$$=\mathrm{1}−\mathrm{tan}\:{A}×\mathrm{tan}\:{B} \\ $$$$\boldsymbol{{P}}=\mathrm{1}+\mathrm{1}−\mathrm{tan}{A}×\mathrm{tan}{B}+\mathrm{tan}\:{A}×\mathrm{tan}\:{B}\:=\mathrm{2} \\ $$$$\:\:{donc} \\ $$$$\left(\mathrm{1}+\mathrm{tan}\:{A}\right)\left(\mathrm{1}+\mathrm{tan}\:{B}\right)=\mathrm{2} \\ $$$$\: \\ $$

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