Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 18184 by Tinkutara last updated on 16/Jul/17

(m+2)sinθ + (2m−1)cosθ = 2m+1, if  (1) tanθ = (3/4)  (2) tanθ = (4/3)  (3) tanθ = ((2m)/(m^2  − 1))  (4) tanθ = ((2m)/(m^2  + 1))

$$\left({m}+\mathrm{2}\right)\mathrm{sin}\theta\:+\:\left(\mathrm{2}{m}−\mathrm{1}\right)\mathrm{cos}\theta\:=\:\mathrm{2}{m}+\mathrm{1},\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}{m}}{{m}^{\mathrm{2}} \:−\:\mathrm{1}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}\theta\:=\:\frac{\mathrm{2}{m}}{{m}^{\mathrm{2}} \:+\:\mathrm{1}} \\ $$

Commented by prakash jain last updated on 16/Jul/17

tan (θ/2)=u  sin θ=((2u)/(1+u^2 )), cos θ=((1−u^2 )/(1+u^2 ))  (m+2)2u+(2m−1)(1−u^2 )=(2m+1)(1+u^2 )  4mu^2 +2(m+2)u+2=0  2mu^2 −(m+2)u+1=0  (2u−1)(mu−1)=0  u=(1/2),(1/m)  tan θ=((2tan θ/2)/(1−tan^2 θ/2))  tan θ=(4/3),((2m)/(m^2 −1))

$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}={u} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} },\:\mathrm{cos}\:\theta=\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\left({m}+\mathrm{2}\right)\mathrm{2}{u}+\left(\mathrm{2}{m}−\mathrm{1}\right)\left(\mathrm{1}−{u}^{\mathrm{2}} \right)=\left(\mathrm{2}{m}+\mathrm{1}\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right) \\ $$$$\mathrm{4}{mu}^{\mathrm{2}} +\mathrm{2}\left({m}+\mathrm{2}\right){u}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}{mu}^{\mathrm{2}} −\left({m}+\mathrm{2}\right){u}+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{2}{u}−\mathrm{1}\right)\left({mu}−\mathrm{1}\right)=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{{m}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2tan}\:\theta/\mathrm{2}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \theta/\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{2}{m}}{{m}^{\mathrm{2}} −\mathrm{1}} \\ $$

Commented by Tinkutara last updated on 17/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by ajfour last updated on 16/Jul/17

(√((m+2)^2 +(2m−1)^2 ))sin (θ+tan^(−1) ((2m−1)/(m+2)))=2m+1  θ=−tan^(−1) ((2m−1)/(m+2))+sin^(−1) ((2m+1)/(√((m−2)^2 +(2m+1)^2 )))    =−tan^(−1) ((2m−1)/(m+2))+tan^(−1) ((2m+1)/(m−2))  tan θ=((((2m+1)/(m−2))−((2m−1)/(m+2)))/(1+((2m+1)/(m−2)).((2m−1)/(m+2))))    = (((2m+1)(m+2)−(2m−1)(m−2))/(m^2 −4+4m^2 −1))   tan θ=((10m)/(5(m^2 −1))) =((2m)/(m^2 −1)) .      [ option  (3) ,(2) ]     for  m=2 ,   tan θ=(4/3).

$$\sqrt{\left(\mathrm{m}+\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2m}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{sin}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2m}−\mathrm{1}}{\mathrm{m}+\mathrm{2}}\right)=\mathrm{2m}+\mathrm{1} \\ $$$$\theta=−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2m}−\mathrm{1}}{\mathrm{m}+\mathrm{2}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2m}+\mathrm{1}}{\sqrt{\left(\mathrm{m}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{2m}+\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$\:\:=−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2m}−\mathrm{1}}{\mathrm{m}+\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2m}+\mathrm{1}}{\mathrm{m}−\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=\frac{\frac{\mathrm{2m}+\mathrm{1}}{\mathrm{m}−\mathrm{2}}−\frac{\mathrm{2m}−\mathrm{1}}{\mathrm{m}+\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{2m}+\mathrm{1}}{\mathrm{m}−\mathrm{2}}.\frac{\mathrm{2m}−\mathrm{1}}{\mathrm{m}+\mathrm{2}}} \\ $$$$\:\:=\:\frac{\left(\mathrm{2m}+\mathrm{1}\right)\left(\mathrm{m}+\mathrm{2}\right)−\left(\mathrm{2m}−\mathrm{1}\right)\left(\mathrm{m}−\mathrm{2}\right)}{\mathrm{m}^{\mathrm{2}} −\mathrm{4}+\mathrm{4m}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:\mathrm{tan}\:\theta=\frac{\mathrm{10m}}{\mathrm{5}\left(\mathrm{m}^{\mathrm{2}} −\mathrm{1}\right)}\:=\frac{\mathrm{2m}}{\mathrm{m}^{\mathrm{2}} −\mathrm{1}}\:. \\ $$$$\:\:\:\:\left[\:\mathrm{option}\:\:\left(\mathrm{3}\right)\:,\left(\mathrm{2}\right)\:\right] \\ $$$$\:\:\:\mathrm{for}\:\:\mathrm{m}=\mathrm{2}\:,\:\:\:\mathrm{tan}\:\theta=\frac{\mathrm{4}}{\mathrm{3}}. \\ $$

Commented by ajfour last updated on 16/Jul/17

you can view solution of Q.18111  (chord AB at ∠=φ with diameter  cut in the ratio p:q )

$$\mathrm{you}\:\mathrm{can}\:\mathrm{view}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{Q}.\mathrm{18111} \\ $$$$\left(\mathrm{chord}\:\mathrm{AB}\:\mathrm{at}\:\angle=\phi\:\mathrm{with}\:\mathrm{diameter}\right. \\ $$$$\left.\mathrm{cut}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{p}:\mathrm{q}\:\right) \\ $$

Commented by Tinkutara last updated on 16/Jul/17

I viewed it. Nice method! And thanks  for this solution!

$$\mathrm{I}\:\mathrm{viewed}\:\mathrm{it}.\:\mathrm{Nice}\:\mathrm{method}!\:\mathrm{And}\:\mathrm{thanks} \\ $$$$\mathrm{for}\:\mathrm{this}\:\mathrm{solution}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com