What does #cos(arctan(2))-sin(arcsec(5))# equal?

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Answer 1 · 2016-05-19T10:04:26

#cos(arctan2)-sin(arcsec5)# is #+-1.427# or #+-0.5326#

#cos(arctan2)-sin(arcsec5)#

= #cosalpha-sinbeta#, where

#tanalpha=2# and #secbeta=5#

Now, #cosalpha=i/secalpha=1/sqrt(1+tan^2alpha)#

= #1/sqrt(1+2^2)=+-1/sqrt5#

and #sinbeta=sqrt(1-cos^2beta)=sqrt(1-1/sec^2beta)#

= #sqrt(1-1/5^2)=sqrt(1-1/25)=sqrt(24/25)=+-2/5sqrt6#

Hence #cosalpha-sinbeta=(+-1/sqrt5)-(+-2/5sqrt6)#

= #(+-0.4472)-(+-0.9798)#

Taking different sign combinations can take four different values of #cosalpha-sinbeta#, which are #+-1.427# or #+-0.5326#