Question

In a right ∆ABC, right angled at B, the ratio of AB to AC is 1:√2. Find the values of i.2tanA/1+tan2A   ii. sinA cos C+cosA sinC

Answer 1

AB/BC= (1/√2)
that means, base/hypo = (1/√2)
so, cosA= (1/√2)
i.e., angleA= 45°
therefore, angleC = 180-(90+45) = 45°( since angleB= 90°)
so, tan2A= tan90°= infinity
so 2tanA/(1+tan2A)= Infinity

sinAcosC + cosAsinC
=sin(A+C)
=sin90°
=1