Given that tanx=1,0°<=x<=90°

Question

Evaluate 1-sin²x/cosx

Answer 1

Assuming, {1-(sinx)^2}/cosx

{1-(sinx)^2}/cosx

or {(cosx)^2}/cosx , since 1-(sinx)^2 = (cosx)^2

or cosx  -----------(1)

Given, tanx =1 so perpendicular = 1, base=1 so hypotenuse = sqrt( Perpendicular^2 + base^2) = sqrt(1+1) = sqrt(2)

So, from (1) we know,

cosx = base/hypotenuse

cosx = 1/sqrt(2)

1/sqrt(2) is the solution