How do you find the exact value using the sum or difference identities of secant 105 degrees?

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Answer 1 · 2016-01-28T11:13:00

#sec 105^@=-sqrt2-sqrt6#

There is a solution using special angles #45^@# and #60^@#.

#sec 105^@=1/cos (105^@)=1/(cos(60^@+45^@)#

#=1/(cos 60^@ cos 45^@-sin 60^@ sin 45^@)#

#=1/(1/2* 1/sqrt2-sqrt3/2 *1/sqrt2)#

#=1/((1-sqrt3)/(2sqrt2))#

#=(2sqrt2)/(1-sqrt3)#

#=(2sqrt2(1+sqrt3))/(1^2-(sqrt3)^2)#

#=(2sqrt2(1+sqrt3))/(1-3)#

#=(2sqrt2(1+sqrt3))/-2#

#=(cancel2sqrt2(1+sqrt3))/cancel(-2)#

#=-sqrt2(1+sqrt3)#

#=-sqrt2-sqrt6#

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