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Question-90904




Question Number 90904 by  M±th+et+s last updated on 26/Apr/20
Commented by  M±th+et+s last updated on 26/Apr/20
For 0<x<(π/2) , tangents to graphs of  y=cos(x) and y=tan(x) are exteded to  meet the x−axis from the point of   intersection.  find a:b .
$${For}\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}\:,\:{tangents}\:{to}\:{graphs}\:{of} \\ $$$${y}={cos}\left({x}\right)\:{and}\:{y}={tan}\left({x}\right)\:{are}\:{exteded}\:{to} \\ $$$${meet}\:{the}\:{x}−{axis}\:{from}\:{the}\:{point}\:{of}\: \\ $$$${intersection}. \\ $$$${find}\:{a}:{b}\:. \\ $$
Answered by Kunal12588 last updated on 26/Apr/20
let intersection point be (p,q)  tan p = cos p  sin^2 p+sinp−1=0  sin p = ((−1±(√(1+4)))/2)=(((√5)−1)/2)  p = sin^(−1)  (((√5)−1)/2)=tan^(−1) (√(((√5)−1)/2))=cos^(−1) (√(((√5)−1)/2))  m_1 = sec^2 (cos^(−1) (√(((√5)−1)/2)))=(2/( (√5)−1))=(((√5)+1)/2)  m_2 = −sin(sin^(−1) (((√5)−1)/2))=−(((√5)−1)/2)  ∵ m_1 ∙m_2 =−1  ∴ (a/b)=m_1 =(((√5)+1)/2)=1.61803398...
$${let}\:{intersection}\:{point}\:{be}\:\left({p},{q}\right) \\ $$$${tan}\:{p}\:=\:{cos}\:{p} \\ $$$${sin}^{\mathrm{2}} {p}+{sinp}−\mathrm{1}=\mathrm{0} \\ $$$${sin}\:{p}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$${p}\:=\:{sin}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}={tan}^{−\mathrm{1}} \sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}={cos}^{−\mathrm{1}} \sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}} \\ $$$${m}_{\mathrm{1}} =\:{sec}^{\mathrm{2}} \left({cos}^{−\mathrm{1}} \sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$${m}_{\mathrm{2}} =\:−{sin}\left({sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)=−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\because\:{m}_{\mathrm{1}} \centerdot{m}_{\mathrm{2}} =−\mathrm{1} \\ $$$$\therefore\:\frac{{a}}{{b}}={m}_{\mathrm{1}} =\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}=\mathrm{1}.\mathrm{61803398}… \\ $$
Commented by  M±th+et+s last updated on 26/Apr/20
thank you sir and its correct solution
$${thank}\:{you}\:{sir}\:{and}\:{its}\:{correct}\:{solution} \\ $$

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