Question

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

Answer 1

Step-by-step explanation:

we have,

√{(1+sin2A)/(1-sin2A)} = tan(π/4+A)

LHS,,

by rationalizing

= √{(1+sin2A)²/(1-sin²2A)}

= √{(1+sin2A)²/cos²2A}

= √{[1-cos(π/2+2A)]²/sin²(π/2+2A)}

= {[1-cos(π/2+2A)]/[sin(π/2+2A)]}

={2sin²(π/4+A)/2sin(π/4+A)cos(π/4+A)}

= {sin(π/4+A)/cos(π/4+A)}

=> tan(π/4+A) = RHS

hence proved....