If $tan(5x - 10^{0})$ = $cot(5y + 20^{0})$, the value of (x +y ) is
1). $15^{0}$
2). $16^{0}$
3). $24^{0}$
4). $20^{0}$
tan (5x – 10°) = cot (5y + 20°)
=> tan (5x – 10°) = tan (90° –(5y + 20°))
=> 5x – 10° = 90° – 5y – 20°
=> 5x + 5y =70° + 10°
=> 5 (x + y) = 80°
=> x + y = 16° .
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