If sina =3÷5 and cos b=5÷13 find value of sinb + cos a
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(6y - 11)(6y + 11) = ay² - b |use (a - b)(a + b) = a² - b²
(6y)² - 11² =ay² - b
36y² - 121 = ay² - b |add b to both sides
ay² = 36y² - 121 + b |divide both sides by y² ≠ 0
a = (36y² - 121 + b)/y²
(6y)² - 11² =ay² - b
36y² - 121 = ay² - b |add b to both sides
ay² = 36y² - 121 + b |divide both sides by y² ≠ 0
a = (36y² - 121 + b)/y²