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Question Number 49061 by Pk1167156@gmail.com last updated on 02/Dec/18
The numerical value of tan (2 tan^(−1) (1/5) − (π/4)) is
$$\mathrm{The}\:\mathrm{numerical}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\mathrm{tan}\:\left(\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\:−\:\frac{\pi}{\mathrm{4}}\right)\:\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18
tan^(−1) ((1/5))+tan^(−1) ((1/5))=tan^(−1) ((((1/5)+(1/5))/(1−(1/5)×(1/5))))=tan^(−1) (((2/5)/(1−(1/(25))))) =tan^(−1) ((2/5)×((25)/(24)))=tan^(−1) ((5/(12))) tan(tan^(−1) ((5/(12)))−(π/4))=((tan(tan^(−1) ((5/(12))))−tan((π/4)))/(1+tan(tan^(−1) ((5/(12))))tan((π/4)))) =(((5/(12))−1)/(1+(5/(12 ))×1))=((−7)/(17)) pls check...
$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)={tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}×\frac{\mathrm{1}}{\mathrm{5}}}\right)={tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{2}}{\mathrm{5}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{5}}×\frac{\mathrm{25}}{\mathrm{24}}\right)={tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{12}}\right) \\ $$$${tan}\left({tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{12}}\right)−\frac{\pi}{\mathrm{4}}\right)=\frac{{tan}\left({tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{12}}\right)\right)−{tan}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{1}+{tan}\left({tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{12}}\right)\right){tan}\left(\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\frac{\frac{\mathrm{5}}{\mathrm{12}}−\mathrm{1}}{\mathrm{1}+\frac{\mathrm{5}}{\mathrm{12}\:}×\mathrm{1}}=\frac{−\mathrm{7}}{\mathrm{17}}\:{pls}\:{check}… \\ $$
Answered by hknkrc46 last updated on 02/Dec/18
tan (2tan^(−1) (1/5)−(𝛑/4))=((tan (2tan^(−1) (1/5))−tan (𝛑/4))/(1+tan (2tan^(−1) (1/5))tan (𝛑/4))) =((tan (tan^(−1) (1/5)+tan^(−1) (1/5))−tan (𝛑/4))/(1+tan (tan^(−1) (1/5)+tan^(−1) (1/5))tan (𝛑/4))) =((((tan (tan^(−1) (1/5))+tan (tan^(−1) (1/5)))/(1−tan (tan^(−1) (1/5))tan (tan^(−1) (1/5))))−tan (𝛑/4))/(1+((tan (tan^(−1) (1/5))+tan (tan^(−1) (1/5)))/(1−tan (tan^(−1) (1/5))tan (tan^(−1) (1/5))))tan (𝛑/4))) =(((((1/5)+(1/5))/(1−(1/5)∙(1/5)))−1)/(1+(((1/5)+(1/5))/(1−(1/5)∙(1/5)))))=((((2/5)/((24)/(25)))−1)/(1+((2/5)/((24)/(25)))))=(((5/(12))−1)/(1+(5/(12))))=((−(7/(12)))/((17)/(12)))=−(7/(17)) ★ {tan (tan^(−1) (1/5))=(1/5) ; tan (𝛑/4)=1}
$$\mathrm{tan}\:\left(\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}−\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)=\frac{\mathrm{tan}\:\left(\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}}{\mathrm{1}+\mathrm{tan}\:\left(\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}}{\mathrm{1}+\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} \\ $$$$=\frac{\frac{\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)}{\mathrm{1}−\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)}−\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}}{\mathrm{1}+\frac{\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)}{\mathrm{1}−\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)}\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} \\ $$$$=\frac{\frac{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\centerdot\frac{\mathrm{1}}{\mathrm{5}}}−\mathrm{1}}{\mathrm{1}+\frac{\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\centerdot\frac{\mathrm{1}}{\mathrm{5}}}}=\frac{\frac{\frac{\mathrm{2}}{\mathrm{5}}}{\frac{\mathrm{24}}{\mathrm{25}}}−\mathrm{1}}{\mathrm{1}+\frac{\frac{\mathrm{2}}{\mathrm{5}}}{\frac{\mathrm{24}}{\mathrm{25}}}}=\frac{\frac{\mathrm{5}}{\mathrm{12}}−\mathrm{1}}{\mathrm{1}+\frac{\mathrm{5}}{\mathrm{12}}}=\frac{−\frac{\mathrm{7}}{\mathrm{12}}}{\frac{\mathrm{17}}{\mathrm{12}}}=−\frac{\mathrm{7}}{\mathrm{17}} \\ $$$$\bigstar\:\left\{\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{5}}\right)=\frac{\mathrm{1}}{\mathrm{5}}\:;\:\mathrm{tan}\:\frac{\boldsymbol{\pi}}{\mathrm{4}}=\mathrm{1}\right\} \\ $$

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