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**We know that, If secA+tanA=p, then secA-tanA=1/p**

**Given,**

**secA+tanA=5**

**Then secA-tanA=1/5**

**Case 1:**

**By adding 1 and 2 we get,**

**2secA=5+1/5 (since, +tanA and -tanA gets cancelled)**

**2secA = 25+1/5 (LCM)**

**secA = 26/5 × 1/2**

**secA = 13/5**

**secA will be positive on Q1 and Q4 ------- 1`**

**Case 2:**

**By subtracting 1 and 2**

**2tanA = 5-1/5 (since, +secA and -secA gets cancelled)**

**2tanA = 25-1/5 (LCM)**

**tanA = 24/5 × 1/2**

**tanA = 12/5**

**tanA will be positive on Q1 and Q3 ------- 2**

from 1 and 2 markings, Q1 is common

from 1 and 2 markings, Q1 is common

**so, A belongs to Q1**

**From calculation, we get tanA = 12/5 = opp side/ adj side**

**Then hypotenuse = 13 (by pythagoras theorem)**

**Then, sinA = opp side/hypotenuse = 12/13**

**Therefore, sinA = 12/13**