Question

show that root 1+sin2A ÷ 1- sin 2A = tan (pie ÷4 +A )​

Answer 1

Explanation:

$(1 + sin2A) \div (1 - sin2A) = tan( \frac{ \pi}{4} + A) \\ LHS = (1 + sin2A) \div (1 - sin2A) \\ = \frac{1 + sin2A}{1 - sin2A} \\ = \frac{ {sin}^{2} A + {cos}^{2} A + 2sinA cosA}{{sin}^{2} A + {cos}^{2} A - 2sinA cosA} \\ = \frac{ cosA + sinA}{cosA - sinA} \\ dividing \: numerator \: and \: denominator \: \\ by \: cosA \: \\ = \frac{ 1 + tanA}{1 - tanA} \\ = \frac{ tan \frac{ \pi}{4} + tanA}{1 - tan \frac{ \pi}{4} tanA} \: \\ = tan (\frac{ \pi}{4} + A) \\ =RHS \\ thus \: proved.$