How do you solve #4\sec ^{2}\Theta =5# ?

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Answer 1 · 2017-11-04T09:05:51

#theta = 26.5# degree

#sec^2 theta = 5/4#
#cos^2theta = 4/5#

Now

#(1+cos2theta)/2 = cos^2theta#

Therefore,

#(1+cos2theta)/2 = 4/5#
#(1+cos2theta) = 8/5#

Rearranging it

#cos2theta=3/5#

therefore

#cos2theta=cos53# degree

that means

#2theta = 53# degree

#theta = 26.5# degree

Answer 2 · 2017-11-04T12:33:19

#26.57^o#

Using identity:

#sec^2theta= 1 + tan^2theta#

#4(1+tan^2theta)=5#

#4+4tan^2theta=5#

#tan^2theta=1/4#

#tantheta= sqrt(1/4)=1/2#

#theta= arctan(tantheta) = arctan(1/2)=26.57^o# ( 2 .d.p )