Question-13224

Question Number 13224 by Tinkutara last updated on 16/May/17

Commented by Tinku Tara last updated on 16/May/17

Can u please use a different name for your user id. Since tinku tara is publisher name. Thank You for ur cooperation.

$$\mathrm{Can}\:\mathrm{u}\:\mathrm{please}\:\mathrm{use}\:\mathrm{a}\:\mathrm{different}\:\mathrm{name} \\ $$$$\mathrm{for}\:\mathrm{your}\:\mathrm{user}\:\mathrm{id}.\:\mathrm{Since}\:\mathrm{tinku}\:\mathrm{tara} \\ $$$$\mathrm{is}\:\mathrm{publisher}\:\mathrm{name}. \\ $$$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{for}\:\mathrm{ur}\:\mathrm{cooperation}. \\ $$

Commented by Tinku Tara last updated on 17/May/17

To create a new username. Reinstall this app amd create a different user id.

$$\mathrm{To}\:\mathrm{create}\:\mathrm{a}\:\mathrm{new}\:\mathrm{username}.\:\mathrm{Reinstall} \\ $$$$\mathrm{this}\:\mathrm{app}\:\mathrm{amd}\:\mathrm{create}\:\mathrm{a}\:\mathrm{different}\:\mathrm{user}\:\mathrm{id}. \\ $$

Commented by Tinkutara last updated on 17/May/17

If I reinstall this app, then all the question ID data will be lost. What is the problem with this username?

$$\mathrm{If}\:\mathrm{I}\:\mathrm{reinstall}\:\mathrm{this}\:\mathrm{app},\:\mathrm{then}\:\mathrm{all}\:\mathrm{the} \\ $$$$\mathrm{question}\:\mathrm{ID}\:\mathrm{data}\:\mathrm{will}\:\mathrm{be}\:\mathrm{lost}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{problem}\:\mathrm{with}\:\mathrm{this}\:\mathrm{username}? \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17

α+β=π+γ⇒sin(α+β)=sin(π+γ)=−sinγ sin^2 (α+β)=(−sinγ)^2 sin^2 α.cos^2 β+2sinαcosαsinβcosβ+ +cos^2 αsin^2 β=sin^2 γ sin^2 α(1−sin^2 β)+sin^2 β(1−sin^2 α)+ +2sinαcosαsinβcosβ=sin^2 γ⇒ sin^2 α+sin^2 β−sin^2 γ=2sinαsinβ(sinαsinβ−cosαcosβ)= =−2sinαsinβcos(α+β)= =−2sinαsinβcos(π+γ)= =2sin𝛂sin𝛃cos𝛄⇒𝛌=2 .■

$$\alpha+\beta=\pi+\gamma\Rightarrow{sin}\left(\alpha+\beta\right)={sin}\left(\pi+\gamma\right)=−{sin}\gamma \\ $$$${sin}^{\mathrm{2}} \left(\alpha+\beta\right)=\left(−{sin}\gamma\right)^{\mathrm{2}} \\ $$$${sin}^{\mathrm{2}} \alpha.{cos}^{\mathrm{2}} \beta+\mathrm{2}{sin}\alpha{cos}\alpha{sin}\beta{cos}\beta+ \\ $$$$+{cos}^{\mathrm{2}} \alpha{sin}^{\mathrm{2}} \beta={sin}^{\mathrm{2}} \gamma \\ $$$${sin}^{\mathrm{2}} \alpha\left(\mathrm{1}−{sin}^{\mathrm{2}} \beta\right)+{sin}^{\mathrm{2}} \beta\left(\mathrm{1}−{sin}^{\mathrm{2}} \alpha\right)+ \\ $$$$+\mathrm{2}{sin}\alpha{cos}\alpha{sin}\beta{cos}\beta={sin}^{\mathrm{2}} \gamma\Rightarrow \\ $$$${sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta−{sin}^{\mathrm{2}} \gamma=\mathrm{2}{sin}\alpha{sin}\beta\left({sin}\alpha{sin}\beta−{cos}\alpha{cos}\beta\right)= \\ $$$$\:\:\:=−\mathrm{2}{sin}\alpha{sin}\beta{cos}\left(\alpha+\beta\right)= \\ $$$$\:\:\:=−\mathrm{2}{sin}\alpha{sin}\beta{cos}\left(\pi+\gamma\right)= \\ $$$$\:\:\:=\mathrm{2}\boldsymbol{{sin}\alpha{sin}\beta{cos}\gamma}\Rightarrow\boldsymbol{\lambda}=\mathrm{2}\:\:\:.\blacksquare \\ $$$$ \\ $$

Commented by Tinkutara last updated on 17/May/17

$$\mathrm{Thanks}\:\mathrm{a}\:\mathrm{lot}! \\ $$

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