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Question Number 43642 by gunawan last updated on 13/Sep/18

$$\mathrm{The}\mathrm{most}\mathrm{general}\mathrm{value}\mathrm{of}\theta \mathrm{which}$$$$\mathrm{satisfy}\mathrm{both}\mathrm{the}\mathrm{equations}\cos \theta =-\frac{1}{\sqrt{2}}$$$$\mathrm{and}\tan \theta =1\mathrm{is}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Sep/18

$$cos\theta =-vetan\theta =+veso\theta liesinthirdquadrant$$$$tan\theta =1=tan(\mathrm{\Pi }+\frac{\mathrm{\Pi }}{4})\theta =\frac{5\mathrm{\Pi }}{4}$$$$check...$$$$cos(\mathrm{\Pi }+\frac{\mathrm{\Pi }}{4})=-cos\left(\frac{\mathrm{\Pi }}{4}\right)=-\frac{1}{\sqrt{2}}$$$$tan\theta =tan\frac{5\mathrm{\Pi }}{4}$$$$\theta =x\mathrm{\Pi }+\frac{5\mathrm{\Pi }}{4}x=2\theta =\frac{13\mathrm{\Pi }}{4}$$$$$$$$cos\theta =cos\left(\frac{5\mathrm{\Pi }}{4}\right)$$$$\theta =2y\mathrm{\Pi }\pm \frac{5\mathrm{\Pi }}{4}$$$$\theta =2y\mathrm{\Pi }+\frac{5\mathrm{\Pi }}{4}y=1\theta =\frac{13\mathrm{\Pi }}{4}$$$$$$$$$$$$$$

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