How to solve #tan theta + 3 cot theta = 5 sec theta# in quadratic form?

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Answer 1 · 2018-06-14T08:56:23

#tantheta+3cottheta=5sectheta#

#=>sintheta/costheta+3costheta/sintheta=5/costheta#

#=>sin^2theta+3cos^2theta=5sintheta#

#=>sin^2theta+3-3sin^2theta=5sintheta#

#=>2sin^2theta+5sintheta-3=0#

#=>2sin^2theta+6sintheta-sintheta-3=0#

#=>2sintheta(sintheta+3)-(sintheta+3)=0#

#=>(sintheta+3)(2sintheta-1)=0#

Since #-1<=sintheta<=1# so #sintheta=-3# not possible.

Hence #sintheta=1/2=sin(pi/6)#

So #theta=npi+(-1)^npi/6" where " n inZZ#