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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

$$For0<x<\frac{\pi }{2},tangentstographsof$$$$y=cos\left(x\right)andy=tan\left(x\right)areextededto$$$$meetthex-axisfromthepointof$$$$intersection.$$$$finda:b.$$

Answered by Kunal12588 last updated on 26/Apr/20

$$letintersectionpointbe(p,q)$$$$tanp=cosp$$$${sin}^{2}p+sinp-1=0$$$$sinp=\frac{-1\pm \sqrt{1+4}}{2}=\frac{\sqrt{5}-1}{2}$$$$p={sin}^{-1}\frac{\sqrt{5}-1}{2}={tan}^{-1}\sqrt{\frac{\sqrt{5}-1}{2}}={cos}^{-1}\sqrt{\frac{\sqrt{5}-1}{2}}$$$$m_1={sec}^2\left({cos}^{-1}\sqrt{\frac{\sqrt{5}-1}{2}}\right)=\frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2}$$$$m_2=-sin\left({sin}^{-1}\frac{\sqrt{5}-1}{2}\right)=-\frac{\sqrt{5}-1}{2}$$$$\because m_1\cdot m_2=-1$$$$\therefore \frac{a}{b}=m_1=\frac{\sqrt{5}+1}{2}=1.61803398...$$

Commented by  M±th+et+s last updated on 26/Apr/20

$$thankyousiranditscorrectsolution$$

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