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Question Number 154240 by amin96 last updated on 15/Sep/21

{ ((tg(a+b)=7)),((tg(a−b)=5)) :} tg(a)=? easy question

$$\begin{cases}{{tg}\left({a}+{b}\right)=\mathrm{7}}\\{{tg}\left({a}−{b}\right)=\mathrm{5}}\end{cases}\:\:{tg}\left({a}\right)=?\:\:\:{easy}\:{question} \\ $$

Answered by mr W last updated on 15/Sep/21

a+b=tan^(−1) 7 a−b=tan^(−1) 5 2a=tan^(−1) 7+tan^(−1) 5 tan 2a=((7+5)/(1−7×5))=−(6/(17))=((2tan a)/(1−tan^2 a)) 3tan^2 a−17 tan a−3=0 ⇒tan a=((17±5(√(13)))/6)

$${a}+{b}=\mathrm{tan}^{−\mathrm{1}} \mathrm{7} \\ $$$${a}−{b}=\mathrm{tan}^{−\mathrm{1}} \mathrm{5} \\ $$$$\mathrm{2}{a}=\mathrm{tan}^{−\mathrm{1}} \mathrm{7}+\mathrm{tan}^{−\mathrm{1}} \mathrm{5} \\ $$$$\mathrm{tan}\:\mathrm{2}{a}=\frac{\mathrm{7}+\mathrm{5}}{\mathrm{1}−\mathrm{7}×\mathrm{5}}=−\frac{\mathrm{6}}{\mathrm{17}}=\frac{\mathrm{2tan}\:{a}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{a}} \\ $$$$\mathrm{3tan}^{\mathrm{2}} \:{a}−\mathrm{17}\:\mathrm{tan}\:{a}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:{a}=\frac{\mathrm{17}\pm\mathrm{5}\sqrt{\mathrm{13}}}{\mathrm{6}} \\ $$

Commented by SLVR last updated on 26/Sep/21

sir..tan^(−1) 7+tan^(−1) 5 can be π+tan^(−1) (((7+5)/(1−7×5)))..i hope

$${sir}..{tan}^{−\mathrm{1}} \mathrm{7}+{tan}^{−\mathrm{1}} \mathrm{5}\:{can}\:{be} \\ $$$$\pi+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{7}+\mathrm{5}}{\mathrm{1}−\mathrm{7}×\mathrm{5}}\right)..{i}\:{hope} \\ $$

Commented by SLVR last updated on 26/Sep/21

its a condition x>0 y>0and if xy>1itcan be π+tan^(−1) ().

$${its}\:{a}\:{condition}\:{x}>\mathrm{0}\:{y}>\mathrm{0}{and}\: \\ $$$${if}\:{xy}>\mathrm{1}{itcan}\:{be}\:\pi+{tan}^{−\mathrm{1}} \left(\right). \\ $$

Commented by mr W last updated on 26/Sep/21

that doesn′t affect the result, since we take tan (...) at the end. here you can see: a+b=nπ+tan^(−1) 7 a−b=mπ+tan^(−1) 5 2a=nπ+tan^(−1) 7+mπ+tan^(−1) 5 tan 2a=((tan (nπ+tan^(−1) 7)+tan (mπ+tan^(−1) 5))/(1−tan (nπ+tan^(−1) 7)×tan (mπ+tan^(−1) 5))) tan 2a=((7+5)/(1−7×5))=−(6/(17))=((2tan a)/(1−tan^2 a))

$${that}\:{doesn}'{t}\:{affect}\:{the}\:{result}, \\ $$$${since}\:{we}\:{take}\:\mathrm{tan}\:\left(...\right)\:{at}\:{the}\:{end}. \\ $$$${here}\:{you}\:{can}\:{see}: \\ $$$${a}+{b}={n}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{7} \\ $$$${a}−{b}={m}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{5} \\ $$$$\mathrm{2}{a}={n}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{7}+{m}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{5} \\ $$$$\mathrm{tan}\:\mathrm{2}{a}=\frac{\mathrm{tan}\:\left({n}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{7}\right)+\mathrm{tan}\:\left({m}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{5}\right)}{\mathrm{1}−\mathrm{tan}\:\left({n}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{7}\right)×\mathrm{tan}\:\left({m}\pi+\mathrm{tan}^{−\mathrm{1}} \mathrm{5}\right)} \\ $$$$\mathrm{tan}\:\mathrm{2}{a}=\frac{\mathrm{7}+\mathrm{5}}{\mathrm{1}−\mathrm{7}×\mathrm{5}}=−\frac{\mathrm{6}}{\mathrm{17}}=\frac{\mathrm{2tan}\:{a}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{a}} \\ $$

Commented by SLVR last updated on 26/Sep/21

wow...thats right...prof...W

$${wow}...{thats}\:{right}...{prof}...{W} \\ $$$$ \\ $$

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