# Thread: Somebody plz try out proving this indentity....

1. ## Somebody plz try out proving this indentity....

If tanΘ = 4/3,

then prove that:√(1-sin Θ)/(1+sin Θ) = 1/3

2. 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?

3. Originally Posted by Plato
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??

4. Originally Posted by snigdha
umm....nope...
i posted the question as it was given in my book...
what do you think will be the correct identity then??
It is correct if $\displaystyle \theta\in I$. Otherwise no.

5. 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.

6. Originally Posted by blackcompe
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....

7. Hint :

$\displaystyle \sin \theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac{1}{\sec^2\theta}} = \sqrt{1-\frac{1}{1+\tan^2\theta}}$

8. Originally Posted by danielomalmsteen
Hint :

$\displaystyle \sin \theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac{1}{\sec^2\theta}} = \sqrt{1-\frac{1}{1+\tan^2\theta}}$

wow!!
you are genius...!! how did you work this out..?!
i had spent almost 3days on this problem in vain..!!!