Question

1+cotA/cosA + 1+tanA/sinA=2(secA+cosecA)

Answer 1

Step-by-step explanation:

I'll assume this is a problem to prove, and should read

Show 1+tanAsinA+1+cotAcosA=2(secA+cscA)

Let's just get the common denominator and add and see what happens.

1+tanAsinA+1+cotAcosA

=cosA(1+sinAcosA)+sinA(1+cosAsinA)sinAcosA

=cosA+sinA+sinA+cosAsinAcosA

=2cosAsinAcosA+2sinAsinAcosA

=2(1sinA+1cosA)

=2(cscA+secA)

=2(secA+cscA) 

Answer 2

$\frac{1 + cotA}{cosA} + \frac{1 + tanA}{sinA} \\ \\ = \frac{sinA + cosA}{sinA \: cosA} + \frac{cosA + sinA}{sinA \: cosA} \\ \\ = (sinA + cosA)( \frac{2}{sinA \: cosA} )$

$= 2( \frac{sinA + cosA}{sinA \: cosA} ) \\ \\ = 2(secA + cosecA)$