There is difficulity with this problem as stated.
If $\displaystyle \tan(\theta)=\frac{4}{3}$ then $\displaystyle \sin(\theta)=\pm\frac{4}{5}$.
That identity is true for only one of those values.
Did you leave some information out?
There is difficulity with this problem as stated.
If $\displaystyle \tan(\theta)=\frac{4}{3}$ then $\displaystyle \sin(\theta)=\pm\frac{4}{5}$.
That identity is true for only one of those values.
Did you leave some information out?
umm....nope...
i posted the question as it was given in my book...
what do you think will be the correct identity then??
Draw a 3-4-5 triangle where the hypotenuse is 5, the vertical segment is 4 and theta is opposite to it. Let theta = y. Therefore, siny is 4/5 and cosy is 3/5 Since tany = siny/cosy, tany = (4/5)/(3/5) or 4/3.
Draw a 3-4-5 triangle where the hypotenuse is 5, the vertical segment is 4 and theta is opposite to it. Let theta = y. Therefore, siny is 4/5 and cosy is 3/5 Since tany = siny/cosy, tany = (4/5)/(3/5) or 4/3.
But we are required to prove √(1-sin Θ)/(1+sin Θ) = 1/3 tanΘ = 4/3 is the information given to us....