Proving identities?

L.H.S.
= sin²x + cos²x + tan²x
= (sin²x + cos²x) + tan²x
= 1 + tan²x
= sec²x
= (1/cosx)²
= 1/cos²x
= R.H.S.

Hence, sin²x + cos²x + tan²x = 1/cos²x


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L.H.S.
= sinx / (sinx + cosx)
= (sinx / cosx) / [(sinx + cosx) / cosx] ...... Divide both numerator and denominator by cosx
= (sinx / cosx) / [(sinx / cosx) + (cosx / cosx)]
= tanx / (tanx + 1) ...... for sinx / cosx = tanx
= tanx / (1 + tanx)
= R.H.S.

Hence, sinx / (sinx + cosx) = tanx / (1 + tanx)

sin ^2 x + cos ^2 x + tan ^2 x = 1/cos^2 x
1 + tan^2 x = 1/cos^2 x
sec^2 x = 1/cos^2 x
sec^2 x = sec^2 x

it is well know by all who study trig that 1 + tan² Θ = sec² Θ...and if YOU divide the top and bottom of the left side by cos x you will obtain the right side in the 2nd part of this query