MHB Trig Identity Theory: Show $\tan\left({\theta}\right)=\frac{q-p}{2\sqrt{qp}}$

[karush]

If $\sin\left({\theta}\right)=\frac{\left(p-q\right)}{\left(p+q\right)}$

And $p$ and $q$ are $90^o<\theta<180^o$ and $p>q$

Show that $\tan\left({\theta}\right)=\frac{q-p}{2\sqrt{qp}}$

I tried using $q=\frac{2\pi}{3 }$ and $p=\frac{5\pi}{6}$

But not...

To do this but theory I'm clueless

 

[kaliprasad]

If $\sin\left({\theta}\right)=\frac{\left(p-q\right)}{\left(p+q\right)}$

And $p$ and $q$ are $90^o<\theta<180^o$ and $p>q$

Show that $\tan\left({\theta}\right)=\frac{q-p}{2\sqrt{qp}}$

I tried using $q=\frac{2\pi}{3 }$ and $p=\frac{5\pi}{6}$

But not...

To do this but theory I'm clueless

$q=\frac{2\pi}{3 }$ and $p=\frac{5\pi}{6}$ cannot be taken as these I presume you are taking angles but these should be values
you know $\sin \theta$
from this u can evaluate $\cos \theta$
because it is related to square root u get 2 values
because cos is -ve in second qudrant take -ve value
devide and get result

$\sin\left({\theta}\right)=\frac{\left(p-q\right)}{\left(p+q\right)}$

$\cos(\theta) = \sqrt{1-(\frac{\left(p-q\right)}{\left(p+q\right)})^2}$
or $\cos^2(\theta) =\frac{(p+q)^2 -(p-q)^2}{(p-q)^2}= \frac{4pq}{((p-q)^2}$
you should be able to proceed

 

[MarkFL]

Another way to go would be to begin with the Pythagorean identity:

$$\cot^2(\theta)=\csc^2(\theta)-1$$

From this you can solve for $\cot(\theta)$ since you know $$\csc(\theta)\equiv\frac{1}{\sin(\theta)}$$ and bearing in mind which quadrant $\theta$ is in when you take the square root. Then use $$\tan(\theta)\equiv\frac{1}{\cot(\theta)}$$. :)

 

[Greg]

... or you could proceed geometrically: construct a right-angled triangle with opposite side p - q and hypotenuse p + q. Then use the Pythagorean theorem to calculate the adjacent side and recall that the tangent is negative over the given interval.

 

[karush]

The triangle worked for me
$$\sin\left({\theta}\right)=\frac{p-q}{p+q}=\frac{o} {h} $$

Therefore
$$\cos\left({\theta}\right)=-\sqrt{\left(p+q\right)^2 - \left(p-q\right)^2}=-2\sqrt{pq}$$

So (the sign is given by the quadrant)

$$\tan\left({\theta}\right)=\frac{o}{-a}=\frac{q-p}{2\sqrt{qp}}$$

Kinda messy but..

 

[Greg]

$$\cos\theta=\dfrac{2\sqrt{pq}}{p+q}\text{ for }0^\circ<\theta<90^\circ$$, but you don't need that.

$$\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{p-q}{2\sqrt{pq}}\text{ for }0^\circ<\theta<90^\circ$$. For $$90^\circ<\theta<180^\circ$$, $\tan\theta=\dfrac{q-p}{2\sqrt{pq}}$.

I suggest making an effort to understand the other two methods given. :)

 

[karush]

OK, I should of called it adjacent side not $\cos\left({\theta}\right)$

The other suggestions seemed harder..

 

[MarkFL]

OK, I should of called it adjacent side not $\cos\left({\theta}\right)$

The other suggestions seemed harder..

Using the method I suggested, we have:

$$\cot^2(\theta)=\left(\frac{p+q}{p-q}\right)^2-1=\frac{4pq}{(p-q)^2}$$

Since the cotangent function is negative in quadrant II, we take:

$$\cot(\theta)=\frac{2\sqrt{pq}}{q-p}$$

Thus:

$$\tan(\theta)=\frac{q-p}{2\sqrt{pq}}$$

 

[karush]

In method that probably is easier. But it's hard to see what's going on without a triangle.