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Question Number 71242 by TawaTawa last updated on 13/Oct/19

$$\mathrm{Given}:\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{d}}{\mathrm{c}}$$$$\mathrm{Show}\mathrm{that},\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}-\frac{{\mathrm{c}}^2}{{\mathrm{d}}^2}=\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{d}}^2}{{\mathrm{c}}^2}$$

Answered by MJS last updated on 13/Oct/19

$$\alpha +\beta =\frac{1}{\alpha }+\frac{1}{\beta }$$$$\Rightarrow \beta =-\alpha \vee \beta =\frac{1}{\alpha }$$$${\alpha }^2-{\beta }^2=\frac{1}{{\alpha }^2}-\frac{1}{{\beta }^2}\{\begin{array}{l}\beta =-\alpha \Rightarrow \mathrm{true}\mathrm{for}\alpha \ne 0\\ \beta =\frac{1}{\alpha }\Rightarrow \mathrm{true}\mathrm{only}\mathrm{for}\alpha =\pm 1\end{array}$$$$\Rightarrow \mathrm{question}\mathrm{wrong}?{\alpha }^2-{\beta }^2=\frac{1}{{\beta }^2}-\frac{1}{{\alpha }^2}\mathrm{is}\mathrm{always}$$$$\mathrm{true}\mathrm{for}\alpha \ne 0\wedge (\beta =-\alpha \vee \beta =\frac{1}{\alpha })$$

Commented by $@ty@m123 last updated on 13/Oct/19

$$Thanksforthisnew\left(forme\right)concept.$$$$Butthenwhereisthemistake$$$$inmysoution.$$$$Iamrepeatedlycheking$$$$butfailedtofind.$$

Commented by MJS last updated on 13/Oct/19

$$\mathrm{you}\mathrm{have}\mathrm{made}\mathrm{no}\mathrm{mistake}\mathrm{and}\mathrm{your}\mathrm{solution}$$$$\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{mine}:\mathrm{the}\mathrm{question}\mathrm{is}‘‘{\mathrm{wrong}}^{″}$$

Commented by TawaTawa last updated on 14/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by JDamian last updated on 13/Oct/19

$$if\mathit{c}turnsinto-\mathit{c},then$$$$\frac{\mathrm{a}}{\mathrm{b}}-\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{d}}{\mathrm{c}}$$$$$$$$(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{c}}{\mathrm{d}})\times (\frac{\mathrm{a}}{\mathrm{b}}-\frac{\mathrm{c}}{\mathrm{d}})=(\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{d}}{\mathrm{c}})\times (\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{d}}{\mathrm{c}})$$$$$$$$\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}-\frac{{\mathrm{c}}^2}{{\mathrm{d}}^2}=\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{d}}^2}{{\mathrm{c}}^2}$$

Commented by $@ty@m123 last updated on 13/Oct/19

$$Ihaveadoubtonthisprocess.$$$$Ifitwouldbesomethinglike:$$$$a+b=c$$$$Wouldyoustillwrite$$$$‘‘if\mathit{c}turnsinto-\mathit{c},{then}^{″}$$

Commented by TawaTawa last updated on 14/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by $@ty@m123 last updated on 13/Oct/19

$$Solution:$$$$\mathrm{Given}:\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{b}}{\mathrm{a}}+\frac{\mathrm{d}}{\mathrm{c}}$$$$\Rightarrow \frac{ad+bc}{bd}=\frac{bc+ad}{ac}$$$$\Rightarrow bd=ac.....\left(1\right)$$$$Now,$$$$LHS=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}-\frac{{\mathrm{c}}^2}{{\mathrm{d}}^2}$$$$=\frac{a^2{d}^2-b^2{c}^2}{b^2{d}^2}$$$$=\frac{a^2{d}^2-b^2{c}^2}{a^2{c}^2}$$$$=\frac{d^2}{c^2}-\frac{b^2}{a^2}$$$$=RHS$$$$pl.checkyourquestion.$$

Commented by TawaTawa last updated on 14/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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