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