Question

slove that: sinA+√3cosA=1

Answer 1

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Answer 2

SOLUTION:
sin A = 1 - √3 cos A
Sq. on both sides:
(sin A)² = (1 - √3 cos A)²
sin²A = 1 - 2√3 cos A + 3 cos²A
sin²A - 3 cos²A = 1 - 2√3 cos A
(1 - cos²A) - 3 cos²A = 1 - 2√3 cos A
1 - 4 cos² A = 1 - 2√3 cos A
1 - 4 cos² A - 1 + 2√3 cos A = 0
- 4 cos² A + 2√3 cos A = 0
-2 ( 2 cos² A - √3 cos A) = 0
2 cos² A - √3 cos A = 0
cos A ( 2 cos A - √3 ) = 0
       Either
cos A = 0       or        2 cos A - √3 = 0
A = cos⁻¹ 0     or    A = cos⁻¹ (√3 ÷ 2)
A = 90°    or     A = 30°

Verification:
→ Put A = 90°
    sin 90° + √3 cos 90° = 1
               1 = 1 ∴ Proved

→ Put A = 30°
     sin 30° + √3 cos 30° = 1
                   2 ≠ 1 ∴ Not Proved 
HENCE;
       A = 90°