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if-sinA-sinB-1-5-then-6cosA-13cosB-cosA-6cosB-




Question Number 154958 by n0y0n last updated on 23/Sep/21
      if , sinA+sinB=(1/5)    then   ((6cosA+13cosB)/(cosA+6cosB))=?
$$\:\: \\ $$$$\:\:\mathrm{if}\:,\:\mathrm{sinA}+\mathrm{sinB}=\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\:\mathrm{then}\:\:\:\frac{\mathrm{6cosA}+\mathrm{13cosB}}{\mathrm{cosA}+\mathrm{6cosB}}=? \\ $$$$\:\: \\ $$
Commented by prakash jain last updated on 23/Sep/21
No unique answer.  1. put A=π/2   2. put B=π/2
$$\mathrm{No}\:\mathrm{unique}\:\mathrm{answer}. \\ $$$$\mathrm{1}.\:\mathrm{put}\:\mathrm{A}=\pi/\mathrm{2}\: \\ $$$$\mathrm{2}.\:\mathrm{put}\:\mathrm{B}=\pi/\mathrm{2} \\ $$
Commented by n0y0n last updated on 23/Sep/21
sinA+sinB=2 then not 1/5
$$\mathrm{sinA}+\mathrm{sinB}=\mathrm{2}\:\mathrm{then}\:\mathrm{not}\:\mathrm{1}/\mathrm{5} \\ $$
Commented by prakash jain last updated on 23/Sep/21
I meant get 2 solutions  once put A=π/2  second solution put B=π/2  you can get many solution since  you have 2 independent variable  and one equation
$$\mathrm{I}\:\mathrm{meant}\:\mathrm{get}\:\mathrm{2}\:\mathrm{solutions} \\ $$$$\mathrm{once}\:\mathrm{put}\:\mathrm{A}=\pi/\mathrm{2} \\ $$$$\mathrm{second}\:\mathrm{solution}\:\mathrm{put}\:\mathrm{B}=\pi/\mathrm{2} \\ $$$${you}\:{can}\:{get}\:{many}\:{solution}\:{since} \\ $$$${you}\:{have}\:\mathrm{2}\:{independent}\:{variable} \\ $$$${and}\:{one}\:{equation} \\ $$

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