If #8sintheta=4+costheta#, what is the value of #tantheta# ?

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Answer 1 · 2018-07-14T08:43:27

#rarrtanx= 3/4#

Let #theta=x# then

#rarr8sinx=4+cosx#

#rarr(8sinx)/cosx=(4+cosx)/cosx=4/cosx+1#

#rarr8tanx=4secx+1#

#rarr8tanx-1=4secx#

Squaring both sides, we get,

#rarr64tan^2x-16tanx+1=16sec^2x=16(1+tan^2x)#

#rarr64tan^2x-16tanx+1=16+16tan^2x#

#rarr48tan^2x-16tanx-15=0#

#rarr48tan^2x+20tanx-36tanx-15=0#

#rarr4tanx(12tanx+5)-3(12tanx+5)=0#

#rarr(12tanx+5)(4tanx-3)=0#

#rarrtanx=-5/12 or 3/4#

But #x=tan^(-1)(-5/12)# does not satisfy the equation so #tanx=3/4#