Prove-the-tan-cos-1-sin-sin-Z-A-

Question Number 164162 by Zaynal last updated on 15/Jan/22

Prove the; (tan 𝛂 + ((cos 𝛂)/(1 + sin 𝛂))) sin 𝛂 = 𝛂 ^([Z.A])

$$\mathrm{Prove}\:\mathrm{the}; \\ $$$$\left(\boldsymbol{{tan}}\:\boldsymbol{\alpha}\:+\:\frac{\boldsymbol{{cos}}\:\boldsymbol{\alpha}}{\mathrm{1}\:+\:\boldsymbol{{sin}}\:\boldsymbol{\alpha}}\right)\:\boldsymbol{{sin}}\:\boldsymbol{\alpha}\:=\:\boldsymbol{\alpha} \\ $$$$\:^{\left[\mathrm{Z}.\mathrm{A}\right]} \\ $$

Commented by som(math1967) last updated on 15/Jan/22

{tan𝛂 +((cos𝛂(1−sin𝛂))/(1−sin^2 𝛂))}×sinα ={tan𝛂+((cos𝛂(1−sin𝛂))/(cos^2 𝛂))}×sin𝛂 ={tan𝛂+(1/(cos𝛂)) −tan𝛂}×sin𝛂 =sec𝛂×sin𝛂=tan𝛂 so (tan𝛂+((cos𝛂)/(1+sin𝛂)))sin𝛂=tan𝛂 ≠𝛂

$$\left\{\boldsymbol{{tan}\alpha}\:+\frac{\boldsymbol{{cos}\alpha}\left(\mathrm{1}−\boldsymbol{{sin}\alpha}\right)}{\mathrm{1}−\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{\alpha}}\right\}×{sin}\alpha \\ $$$$=\left\{\boldsymbol{{tan}\alpha}+\frac{\boldsymbol{{cos}\alpha}\left(\mathrm{1}−\boldsymbol{{sin}\alpha}\right)}{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{\alpha}}\right\}×\boldsymbol{{sin}\alpha} \\ $$$$=\left\{\boldsymbol{{tan}\alpha}+\frac{\mathrm{1}}{\boldsymbol{{cos}\alpha}}\:−\boldsymbol{{tan}\alpha}\right\}×\boldsymbol{{sin}\alpha} \\ $$$$=\boldsymbol{{sec}\alpha}×\boldsymbol{{sin}\alpha}=\boldsymbol{{tan}\alpha} \\ $$$$\boldsymbol{{so}}\:\left(\boldsymbol{{tan}\alpha}+\frac{\boldsymbol{{cos}\alpha}}{\mathrm{1}+\boldsymbol{{sin}\alpha}}\right)\boldsymbol{{sin}\alpha}=\boldsymbol{{tan}\alpha} \\ $$$$\:\neq\boldsymbol{\alpha} \\ $$

Commented by Zaynal last updated on 15/Jan/22