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If-sin-m-2-2mn-m-2-2mn-2n-2-then-prove-that-tan-m-2-2mn-2mn-2n-2-

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Question Number 206970 by MATHEMATICSAM last updated on 02/May/24
If sinθ = ((m^2 + 2mn)/(m^2 + 2mn + 2n^2 )) then prove that tanθ = ((m^2 + 2mn)/(2mn + 2n^2 )) .
$$\mathrm{If}\:\mathrm{sin}\theta\:=\:\frac{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}}{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} }\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{tan}\theta\:=\:\frac{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}}{\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} }\:. \\ $$
Answered by Rasheed.Sindhi last updated on 02/May/24
sinθ = ((m^2 + 2mn)/(m^2 + 2mn + 2n^2 )) sin^2 θ =( ((m^2 + 2mn)/(m^2 + 2mn + 2n^2 )))^2 1−sin^2 θ=1− (((m^2 + 2mn)^2 )/((m^2 + 2mn + 2n^2 )^2 )) cos^2 θ=(((m^2 + 2mn + 2n^2 )^2 −(m^2 + 2mn)^2 )/((m^2 + 2mn + 2n^2 )^2 )) tan^2 θ= ((sin^2 θ )/(cos^2 ))= (((m^2 + 2mn)^2 )/((m^2 + 2mn + 2n^2 )^2 )) × (((m^2 + 2mn + 2n^2 )^2 )/((m^2 + 2mn + 2n^2 )^2 −(m^2 + 2mn)^2 )) tan^2 θ=(((m^2 + 2mn)^2 )/({(m^2 + 2mn + 2n^2 )−(m^2 + 2mn)}{(m^2 + 2mn + 2n^2 )+(m^2 + 2mn)})) =(((m^2 + 2mn)^2 )/(2n^2 (2m^2 + 4mn + 2n^2 ))) =(((m^2 + 2mn)^2 )/(4n^2 (m^2 + 2mn + n^2 ))) =(((m^2 + 2mn)^2 )/(4n^2 (m+n)^2 )) tanθ=±((m(m + 2n))/(2n(m+n)))
$$\mathrm{sin}\theta\:=\:\frac{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}}{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$\mathrm{sin}^{\mathrm{2}} \theta\:=\left(\:\frac{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}}{{m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta=\mathrm{1}−\:\frac{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)^{\mathrm{2}} }{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mathrm{cos}^{\mathrm{2}} \theta=\frac{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)^{\mathrm{2}} }{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)^{\mathrm{2}} }\: \\ $$$$\mathrm{tan}^{\mathrm{2}} \theta=\:\frac{\mathrm{sin}^{\mathrm{2}} \theta\:}{\mathrm{cos}^{\mathrm{2}} \:}=\:\frac{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)^{\mathrm{2}} }{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)^{\mathrm{2}} }\:×\:\frac{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)^{\mathrm{2}} } \\ $$$$\mathrm{tan}^{\mathrm{2}} \theta=\frac{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)^{\mathrm{2}} }{\left\{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)−\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)\right\}\left\{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)+\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)\right\}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)^{\mathrm{2}} }{\mathrm{2}{n}^{\mathrm{2}} \left(\mathrm{2}{m}^{\mathrm{2}} \:+\:\mathrm{4}{mn}\:+\:\mathrm{2}{n}^{\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:\:=\frac{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)^{\mathrm{2}} }{\mathrm{4}{n}^{\mathrm{2}} \left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\:+\:{n}^{\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left({m}^{\mathrm{2}} \:+\:\mathrm{2}{mn}\right)^{\mathrm{2}} }{\mathrm{4}{n}^{\mathrm{2}} \left({m}+{n}\right)^{\mathrm{2}} } \\ $$$$\mathrm{tan}\theta=\pm\frac{{m}\left({m}\:+\:\mathrm{2}{n}\right)}{\mathrm{2}{n}\left({m}+{n}\right)} \\ $$
Answered by Frix last updated on 02/May/24
t=(s/( (√(1−s^2 )))) s^ =(u/v) ⇒ t=(u/( (√(v^2 −u^2 )))) u=m^2 +2mn v=m^2 +2mn+2n^2 t=((m(m+2n))/(2∣(m+n)n∣))
$${t}=\frac{{s}}{\:\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }} \\ $$$$\overset{} {{s}}=\frac{{u}}{{v}}\:\Rightarrow\:{t}=\frac{{u}}{\:\sqrt{{v}^{\mathrm{2}} −{u}^{\mathrm{2}} }} \\ $$$${u}={m}^{\mathrm{2}} +\mathrm{2}{mn} \\ $$$${v}={m}^{\mathrm{2}} +\mathrm{2}{mn}+\mathrm{2}{n}^{\mathrm{2}} \\ $$$${t}=\frac{{m}\left({m}+\mathrm{2}{n}\right)}{\mathrm{2}\mid\left({m}+{n}\right){n}\mid} \\ $$

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