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Question-119548

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Question Number 119548 by benjo_mathlover last updated on 25/Oct/20
Commented by benjo_mathlover last updated on 25/Oct/20
If cos A = tan B , cos B = tan C and cos C = tan A , then sin A is equal to
$$\:{If}\:\mathrm{cos}\:{A}\:=\:\mathrm{tan}\:{B}\:,\:\mathrm{cos}\:{B}\:=\:\mathrm{tan}\:{C}\:{and}\: \\ $$$$\mathrm{cos}\:{C}\:=\:\mathrm{tan}\:{A}\:,\:{then}\:\mathrm{sin}\:{A}\:{is}\:{equal}\:{to} \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
cos^2 A=tan^2 B=sec^2 B−1 cos^2 A=(1/(cos^2 B))−1 cos^2 A=(1/(tan^2 C))−1 [given cosB=tanC] cos^2 A=(1/(sec^2 C−1))−1=(1/((1/(cos^2 C))−1))−1=((cos^2 C)/(1−cos^2 C))−1 cos^2 A=((cos^2 C−1+cos^2 C)/(1−cos^2 C))=((2tan^2 A−1)/(1−tan^2 A)) cos^2 A=((2sin^2 A−cos^2 A)/(cos^2 A−sin^2 A)) sinA=k 1−k^2 =((2k^2 −(1−k^2 ))/((1−k^2 )−k^2 ))=((3k^2 −1)/(1−2k^2 )) k^2 =1−((3k^2 −1)/(1−2k^2 ))=((1−2k^2 −3k^2 +1)/(1−2k^2 )) k^2 −2k^4 =2−5k^2 2k^4 −6k^2 +2=0 k^4 −3k^2 +1=0 k^2 =((3±(√(9−4)))/2)=((3±(√5))/2) k^2 =((6±2(√5))/4)=((((√5) ±1)/2))^2 k=sinA=(((√5) −1)/2)
$${cos}^{\mathrm{2}} {A}={tan}^{\mathrm{2}} {B}={sec}^{\mathrm{2}} {B}−\mathrm{1} \\ $$$$\:\:\:\:{cos}^{\mathrm{2}} {A}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {B}}−\mathrm{1} \\ $$$${cos}^{\mathrm{2}} {A}=\frac{\mathrm{1}}{{tan}^{\mathrm{2}} {C}}−\mathrm{1}\:\:\left[{given}\:\:\:{cosB}={tanC}\right] \\ $$$${cos}^{\mathrm{2}} {A}=\frac{\mathrm{1}}{{sec}^{\mathrm{2}} {C}−\mathrm{1}}−\mathrm{1}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {C}}−\mathrm{1}}−\mathrm{1}=\frac{{cos}^{\mathrm{2}} {C}}{\mathrm{1}−{cos}^{\mathrm{2}} {C}}−\mathrm{1} \\ $$$${cos}^{\mathrm{2}} {A}=\frac{{cos}^{\mathrm{2}} {C}−\mathrm{1}+{cos}^{\mathrm{2}} {C}}{\mathrm{1}−{cos}^{\mathrm{2}} {C}}=\frac{\mathrm{2}{tan}^{\mathrm{2}} {A}−\mathrm{1}}{\mathrm{1}−{tan}^{\mathrm{2}} {A}} \\ $$$${cos}^{\mathrm{2}} {A}=\frac{\mathrm{2}{sin}^{\mathrm{2}} {A}−{cos}^{\mathrm{2}} {A}}{{cos}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {A}} \\ $$$${sinA}={k} \\ $$$$\mathrm{1}−{k}^{\mathrm{2}} =\frac{\mathrm{2}{k}^{\mathrm{2}} −\left(\mathrm{1}−{k}^{\mathrm{2}} \right)}{\left(\mathrm{1}−{k}^{\mathrm{2}} \right)−{k}^{\mathrm{2}} }=\frac{\mathrm{3}{k}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}−\mathrm{2}{k}^{\mathrm{2}} } \\ $$$${k}^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{3}{k}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}−\mathrm{2}{k}^{\mathrm{2}} }=\frac{\mathrm{1}−\mathrm{2}{k}^{\mathrm{2}} −\mathrm{3}{k}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}−\mathrm{2}{k}^{\mathrm{2}} } \\ $$$${k}^{\mathrm{2}} −\mathrm{2}{k}^{\mathrm{4}} =\mathrm{2}−\mathrm{5}{k}^{\mathrm{2}} \\ $$$$\mathrm{2}{k}^{\mathrm{4}} −\mathrm{6}{k}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$${k}^{\mathrm{4}} −\mathrm{3}{k}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${k}^{\mathrm{2}} =\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} =\frac{\mathrm{6}\pm\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}=\left(\frac{\sqrt{\mathrm{5}}\:\pm\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${k}={sinA}=\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by bemath last updated on 25/Oct/20
let sin A = k = tan B → { ((tan A=(k/( (√(1−k^2 )))))),((cos B = (1/( (√(1+k^2 )))) = tan C)) :} → cos A cos B = sin B → cos B cos C = sin C → cos A cos C = sin A (1)×(2)×(3) →(cos A cos B cos C )^2 = sin A sin B sin C ((√(1−k^2 )) . (1/( (√(1+k^2 )))). ((√(1+k^2 ))/( (√(2+k^2 )))) )^2 = k.(k/( (√(1+k^2 )))). (1/( (√(2+k^2 )))) → ((1−k^2 )/(2+k^2 )) = (k^2 /( (√((1+k^2 )(2+k^2 ))))) let k^2 = u (√((2+u)(1+u))) = (((2+u)u)/(1−u)) (2+u)(1+u)= ((u^2 (2+u)^2 )/((1−u)^2 )) (2+u){(1+u)(1−u)^2 −u^2 (2+u)} = 0 (2+u){ (1−u^2 )(1−u)−2u^2 −u^3 } = 0 (2+u) { 1−u−u^2 +u^3 −2u^2 −u^3 } = 0 (2+u) { 1−u−3u^2 } = 0 u=−2 ←rejected → 3u^2 +u−1 = 0 ⇒ u = ((−1+(√(13)))/6) = k^2 k = (√(((√(13))−1)/6))
$${let}\:\mathrm{sin}\:{A}\:=\:{k}\:=\:\mathrm{tan}\:{B} \\ $$$$\rightarrow\begin{cases}{\mathrm{tan}\:{A}=\frac{{k}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}}\\{\mathrm{cos}\:{B}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}\:=\:\mathrm{tan}\:{C}}\end{cases} \\ $$$$\rightarrow\:\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:=\:\mathrm{sin}\:{B} \\ $$$$\rightarrow\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}\:=\:\mathrm{sin}\:{C} \\ $$$$\rightarrow\:\mathrm{cos}\:{A}\:\mathrm{cos}\:{C}\:=\:\mathrm{sin}\:{A} \\ $$$$\left(\mathrm{1}\right)×\left(\mathrm{2}\right)×\left(\mathrm{3}\right)\: \\ $$$$\rightarrow\left(\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}\:\right)^{\mathrm{2}} \:=\:\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C} \\ $$$$\left(\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }\:.\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}.\:\frac{\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}+{k}^{\mathrm{2}} }}\:\right)^{\mathrm{2}} =\:{k}.\frac{{k}}{\:\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }}.\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+{k}^{\mathrm{2}} }} \\ $$$$\rightarrow\:\frac{\mathrm{1}−{k}^{\mathrm{2}} }{\mathrm{2}+{k}^{\mathrm{2}} }\:=\:\frac{{k}^{\mathrm{2}} }{\left.\:\sqrt{\left(\mathrm{1}+{k}^{\mathrm{2}} \right)\left(\mathrm{2}+{k}^{\mathrm{2}} \right.}\right)} \\ $$$${let}\:{k}^{\mathrm{2}} \:=\:{u} \\ $$$$\sqrt{\left(\mathrm{2}+{u}\right)\left(\mathrm{1}+{u}\right)}\:=\:\frac{\left(\mathrm{2}+{u}\right){u}}{\mathrm{1}−{u}} \\ $$$$\left(\mathrm{2}+{u}\right)\left(\mathrm{1}+{u}\right)=\:\frac{{u}^{\mathrm{2}} \left(\mathrm{2}+{u}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}+{u}\right)\left\{\left(\mathrm{1}+{u}\right)\left(\mathrm{1}−{u}\right)^{\mathrm{2}} −{u}^{\mathrm{2}} \left(\mathrm{2}+{u}\right)\right\}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}+{u}\right)\left\{\:\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\left(\mathrm{1}−{u}\right)−\mathrm{2}{u}^{\mathrm{2}} −{u}^{\mathrm{3}} \:\right\}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}+{u}\right)\:\left\{\:\mathrm{1}−{u}−{u}^{\mathrm{2}} +{u}^{\mathrm{3}} −\mathrm{2}{u}^{\mathrm{2}} −{u}^{\mathrm{3}} \:\right\}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}+{u}\right)\:\left\{\:\mathrm{1}−{u}−\mathrm{3}{u}^{\mathrm{2}} \:\right\}\:=\:\mathrm{0} \\ $$$${u}=−\mathrm{2}\:\leftarrow{rejected} \\ $$$$\rightarrow\:\mathrm{3}{u}^{\mathrm{2}} +{u}−\mathrm{1}\:=\:\mathrm{0}\: \\ $$$$\Rightarrow\:{u}\:=\:\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{6}}\:=\:{k}^{\mathrm{2}} \\ $$$${k}\:=\:\sqrt{\frac{\sqrt{\mathrm{13}}−\mathrm{1}}{\mathrm{6}}} \\ $$$$ \\ $$$$ \\ $$

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