How to solve this identity? Thank you!

Question

#1/sina + 1/tana = cot(a/2)#

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Answer 1 · 2017-12-10T12:42:52

If we know our trigonometric inverses right, we can simplify as follows:
#1/sin(a)=csc(a)#
#1/tan(a)=cot(a)#

#1/sin(a)+1/tan(a)=csc(a)+cot(a)#

This is just the result of the half angle formula for #cot#:
#cot(theta/2)=csc(theta)+cot(theta)#

So we can simplify to:
#=cot(a/2)#, which is what we wanted to prove.

Answer 2 · 2017-12-10T12:53:50

See explanation.

#L=1/sinalpha+1/tanalpha=1/sinalpha+cosalpha/sinalpha=#

#=(1+cosalpha)/sinalpha=(1+cos2*(alpha/2))/sin(2*alpha/2)=#

#(1+2cos^2(alpha/2)-1)/(2sin(alpha/2)cos(alpha/2))=#

#(2cos^2(alpha/2))/(2sin(alpha/2)cos(alpha/2))=#

#(cos^(cancel(2)}(alpha/2))/(sin(alpha/2)cancel(cos(alpha/2)))=#

#(cos(alpha/2))/(sin(alpha/2))=cot(alpha/2)=R#