Question

Verify/prove the following:

(sinA+tanA)/(1+secA) = sinA
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tanu = 1+sinu-cos^2u/cosu(1+sinu)
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cotu = (2+cscu/secu)-2cosu
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tanAsinA/tanA+sinA = tanA-sinA/tanAsinA
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cotA+cscA/sinA-cotA-cscA = -secA

Answer 1

My general approach is to change all ratios to sines and cosines.
I will do the first one, and one other, you try the rest using the same approach.

(sinA+tanA)/(1+secA) = sinA
LS = (sinA + sinA/cosA)/(1 + 1/cosA)
= ((sinAcosA + sinA)/cosA)/((cosA + 1)/cosA )
= sinA(cosA + 1)/cosA * cosA/(cosA + )
= sinA
= RS

cotu = (2+cscu/secu)-2cosu
the way you typed it, it is not an identity, you must mean
cotu = ((2+cscu)/secu)-2cosu

RS =((2 + 1/sinu)*cosu) - 2cosu
= 2cosu + cosu/sinu - 2cosu
= cosu/sinu
= cotu
= LS