Question

Question attached in photo please help (correct gets brainliest)

Answer 1

Step-by-step explanation:

cosA/sin(90-A)+sinA/cos(90-A)

cosA/cosA+sinA/sinA

1+1

2

Answer 2

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→ Answer :

$\frac{cosA}{sin(90^o-A)}+\frac{sinA}{cos(90^o-A)}=2$

→ Formulas :

      $sin(A-B)=sinA.cosB-cosA.sinB\\\\cos(A-B)=cosA.cosB+sinA.sinB)$

       $sin90^o=1,cos90^o=0.$

→ Step-by-step explanation :

$\frac{cosA}{sin(90^o-A)}+\frac{sinA}{cos(90^o-A)}\\\\=>\frac{cosA}{sin90^o.cosA-cos90.sinA}+\frac{sinA}{cos90.cosA+sin90.sinA}\\\\=>\frac{cosA}{cosA} +\frac{sinA}{sinA} \\\\=>1+1\\\\=>2$

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