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Question Number 13249 by Tinkutara last updated on 17/May/17

If msinθ = nsin(θ + 2α), then tan(θ + α)cotα equal to [Answer given in my book is ((1 − n)/(1 + n))]

$$\mathrm{If}\:{m}\mathrm{sin}\theta\:=\:{n}\mathrm{sin}\left(\theta\:+\:\mathrm{2}\alpha\right),\:\mathrm{then} \\ $$$$\mathrm{tan}\left(\theta\:+\:\alpha\right)\mathrm{cot}\alpha\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left[\mathrm{Answer}\:\mathrm{given}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{is}\:\frac{\mathrm{1}\:−\:{n}}{\mathrm{1}\:+\:{n}}\right] \\ $$

Commented by prakash jain last updated on 17/May/17

tan (θ+α)cot α =((sin (θ+α))/(cos (θ+α))) . ((cos α)/(sin α)) sin Acos B=(1/2)(sin (A+B)+sin (A−B)) =(((sin (θ+2α)+sin θ)/2)/((sin (θ+2α)+sin (−θ))/2)) =((sin (θ+2α)+sin θ)/(sin (θ+2α)−sin θ)) assuming sinθ≠0 =((((sin (θ+2α))/(sin θ))+1)/(((sin (θ+2α))/(sin θ))−1))=((m+n)/(m−n)) The answer does not match with ur book. I will recheck later.

$$\mathrm{tan}\:\left(\theta+\alpha\right)\mathrm{cot}\:\alpha \\ $$$$=\frac{\mathrm{sin}\:\left(\theta+\alpha\right)}{\mathrm{cos}\:\left(\theta+\alpha\right)}\:.\:\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha} \\ $$$$\mathrm{sin}\:{A}\mathrm{cos}\:{B}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\left({A}+{B}\right)+\mathrm{sin}\:\left({A}−{B}\right)\right) \\ $$$$=\frac{\frac{\mathrm{sin}\:\left(\theta+\mathrm{2}\alpha\right)+\mathrm{sin}\:\theta}{\mathrm{2}}}{\frac{\mathrm{sin}\:\left(\theta+\mathrm{2}\alpha\right)+\mathrm{sin}\:\left(−\theta\right)}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{sin}\:\left(\theta+\mathrm{2}\alpha\right)+\mathrm{sin}\:\theta}{\mathrm{sin}\:\left(\theta+\mathrm{2}\alpha\right)−\mathrm{sin}\:\theta} \\ $$$${assuming}\:{sin}\theta\neq\mathrm{0} \\ $$$$=\frac{\frac{\mathrm{sin}\:\left(\theta+\mathrm{2}\alpha\right)}{\mathrm{sin}\:\theta}+\mathrm{1}}{\frac{\mathrm{sin}\:\left(\theta+\mathrm{2}\alpha\right)}{\mathrm{sin}\:\theta}−\mathrm{1}}=\frac{{m}+{n}}{{m}−{n}} \\ $$$$\mathrm{The}\:\mathrm{answer}\:\mathrm{does}\:\mathrm{not}\:\mathrm{match}\:\mathrm{with} \\ $$$$\mathrm{ur}\:\mathrm{book}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{recheck}\:\mathrm{later}. \\ $$

Commented by Tinkutara last updated on 18/May/17

I was reaching on this same solution. Maybe answer in my book is wrong.

$$\mathrm{I}\:\mathrm{was}\:\mathrm{reaching}\:\mathrm{on}\:\mathrm{this}\:\mathrm{same}\:\mathrm{solution}. \\ $$$$\mathrm{Maybe}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{my}\:\mathrm{book}\:\mathrm{is}\:\mathrm{wrong}. \\ $$

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