Question #39bdb

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Answer 1 · 2017-10-01T07:03:59

43/20

Given, #9cot^2theta=18cosectheta-14#

#rArr 9(cosec^2theta -1) - 18 cosectheta+14 = 0#

#rArr 9cosec^2theta -9 -18cosectheta+14 = 0#

#rArr 9cosec^2theta - 18 cosectheta +5 = 0#

#rArr 9cosec^2theta-15cosectheta-3cosectheta+5 = 0#

#rArr 3cosectheta(3cosectheta-5)-1(3cosectheta-5)#

#rArr (3cosectheta-5)(3cosectheta-1) = 0#

#rArr 3cosectheta-5 = 0, 3cosectheta-1 = 0#

#rArr cosectheta = 5/3 and 1/3# [ cosectheta can not be 1/3 as theta should be between 0 < theta< pi/2]

Hence #cosectheta = 5/3 or sintheta = 3/5, costheta = 4/5 and tan theta = 3/4#

S0, #sintheta + costheta+tantheta = 3/5+4/5+3/4= 43/20#