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If-and-are-conected-by-the-relation-2tan-2-tan-2-tan-2-tan-2-tan-2-tan-2-tan-2-tan-2-tan-2-1-then-which-of-these-are-correct-multi-correct-A-sin-2-sin-2-sin-2-1-

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Question Number 24972 by Rishabh#1 last updated on 30/Nov/17
If α,β and γ are conected by the relation 2tan^2 α tan^2 β tan^2 γ+tan^2 α tan^2 β + tan^2 β tan^2 γ+tan^2 γ tan^2 α=1 then which of these are correct(multi correct) (A)sin^2 α+sin^2 β+ sin^2 γ=1 (B)cos^2 α+cos^2 β+cos^2 γ=2 (C)cos2α+ cos2β+ cos2γ=1 (D)cos(α+β) cos(α−β)= cos^2 γ
$$\mathrm{If}\:\alpha,\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{conected}\:\mathrm{by}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\mathrm{2tan}^{\mathrm{2}} \alpha\:\mathrm{tan}^{\mathrm{2}} \beta\:\mathrm{tan}^{\mathrm{2}} \gamma+\mathrm{tan}^{\mathrm{2}} \alpha\:\mathrm{tan}^{\mathrm{2}} \beta\:+ \\ $$$$\:\:\:\mathrm{tan}^{\mathrm{2}} \beta\:\mathrm{tan}^{\mathrm{2}} \gamma+\mathrm{tan}^{\mathrm{2}} \gamma\:\mathrm{tan}^{\mathrm{2}} \alpha=\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{which}\:\mathrm{of}\:\mathrm{these}\:\mathrm{are}\:\mathrm{correct}\left(\mathrm{multi}\:\mathrm{correct}\right) \\ $$$$\left(\mathrm{A}\right)\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{sin}^{\mathrm{2}} \beta+\:\mathrm{sin}^{\mathrm{2}} \gamma=\mathrm{1}\: \\ $$$$\left(\mathrm{B}\right)\mathrm{cos}^{\mathrm{2}} \alpha+\mathrm{cos}^{\mathrm{2}} \beta+\mathrm{cos}^{\mathrm{2}} \gamma=\mathrm{2} \\ $$$$\left(\mathrm{C}\right)\mathrm{cos2}\alpha+\:\mathrm{cos2}\beta+\:\mathrm{cos2}\gamma=\mathrm{1}\: \\ $$$$\left(\mathrm{D}\right)\mathrm{cos}\left(\alpha+\beta\right)\:\mathrm{cos}\left(\alpha−\beta\right)=\:\mathrm{cos}^{\mathrm{2}} \gamma \\ $$
Answered by ajfour last updated on 30/Nov/17
say 2abc+ab+bc+ca−1=0 2(1+a)(1+b)(1+c)−(1+a)(1+b) −(1+b)(1+c)−(1+c)(1+a)=0 ⇒ 2=Σ(1/(1+a)) ⇒ Σcos^2 α =2 (B) and so Σ(1+cos 2α)=4 Σcos 2α =1 (C) and Σ(1−cos 2α)=2 ⇒ Σsin^2 α =1 (A) As cos 2α+cos 2β = 1−cos 2γ 2cos (α+β)cos (α−β)=2cos^2 γ ⇒ (D) So all are correct.
$${say}\:\:\:\:\mathrm{2}{abc}+{ab}+{bc}+{ca}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)−\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right) \\ $$$$\:\:\:−\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)−\left(\mathrm{1}+{c}\right)\left(\mathrm{1}+{a}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}=\Sigma\frac{\mathrm{1}}{\mathrm{1}+{a}} \\ $$$$\Rightarrow\:\Sigma\mathrm{cos}\:^{\mathrm{2}} \alpha\:=\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({B}\right) \\ $$$${and}\:{so}\:\:\:\Sigma\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right)=\mathrm{4} \\ $$$$\:\:\:\:\:\:\Sigma\mathrm{cos}\:\mathrm{2}\alpha\:=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\left({C}\right) \\ $$$$\:\:\:{and}\:\:\:\Sigma\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\alpha\right)=\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:\:\Sigma\mathrm{sin}\:^{\mathrm{2}} \alpha\:=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\left({A}\right) \\ $$$${As}\:\:\:\mathrm{cos}\:\mathrm{2}\alpha+\mathrm{cos}\:\mathrm{2}\beta\:=\:\mathrm{1}−\mathrm{cos}\:\mathrm{2}\gamma \\ $$$$\:\:\:\:\mathrm{2cos}\:\left(\alpha+\beta\right)\mathrm{cos}\:\left(\alpha−\beta\right)=\mathrm{2cos}\:^{\mathrm{2}} \gamma \\ $$$$\Rightarrow\:\:\:\:\left({D}\right)\: \\ $$$$\:\:{So}\:{all}\:{are}\:{correct}. \\ $$
Commented by Rishabh#1 last updated on 30/Nov/17
thanks a lot. I learnt a new technique.
$${thanks}\:{a}\:{lot}.\:\mathrm{I}\:\mathrm{learnt}\:\mathrm{a}\:\mathrm{new}\:\mathrm{technique}. \\ $$

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