1. ## Trig help

Hey guys,
I am confused with these two questions (8 and 9). I have no idea where to begin off.
These questions are where one trigonometric function is given and find another.

It would really be help if you could post how to work out this questions.

Thanks

2. ## Re: Trig help

you can note that

$\sin(\theta+90^\circ)=\sin(\theta)\cos(90^\circ)+ \cos(\theta)\sin(90^\circ)=\cos(\theta)$

and that

$\cos(\theta+90^\circ)= \cos(\theta)\cos(90^\circ)-\sin(\theta)\sin(90^\circ)=-\sin(\theta)$

so

$\tan(\theta + 90^\circ)=\dfrac{\cos(\theta)}{-\sin(\theta)}$

$\theta$ is obtuse so

$\sin(\theta) > 0$ and $\cos(\theta)<0$

so

$\cos(\theta)=-\sqrt{1-sin^2(\theta)}=-\sqrt{1-k^2}$

and thus

$\tan(\theta+90^\circ)=\dfrac{-\sqrt{1-k^2}}{-k}=\dfrac{\sqrt{1-k^2}}{k}$

for (9)

$\sec^2(\theta)=1+\tan^2(\theta)$

$\sec^2(\theta)-1=a^2-\dfrac 1 2 + \dfrac 1 {16a^2}=\left(a-\dfrac 1 {4a}\right)^2$

so

$\tan(\theta)=a - \dfrac 1 {4a}$

$\sec(\theta)+\tan(\theta)=\left(a + \dfrac 1 {4a}\right) + \left(a - \dfrac 1 {4a}\right)=2a$

3. ## Re: Trig help

Originally Posted by romsek
you can note that

$\sin(\theta+90^\circ)=\sin(\theta)\cos(90^\circ)+ \cos(\theta)\sin(90^\circ)=\cos(\theta)$

and that

$\cos(\theta+90^\circ)= \cos(\theta)\cos(90^\circ)-\sin(\theta)\sin(90^\circ)=-\sin(\theta)$

so

$\tan(\theta + 90^\circ)=\dfrac{\cos(\theta)}{-\sin(\theta)}$

$\theta$ is obtuse so

$\sin(\theta) > 0$ and $\cos(\theta)<0$

so

$\cos(\theta)=-\sqrt{1-sin^2(\theta)}=-\sqrt{1-k^2}$

and thus

$\tan(\theta+90^\circ)=\dfrac{-\sqrt{1-k^2}}{-k}=\dfrac{\sqrt{1-k^2}}{k}$

for (9)

$\sec^2(\theta)=1+\tan^2(\theta)$

$\sec^2(\theta)-1=a^2-\dfrac 1 2 + \dfrac 1 {16a^2}=\left(a-\dfrac 1 {4a}\right)^2$

so

$\tan(\theta)=a - \dfrac 1 {4a}$

$\sec(\theta)+\tan(\theta)=\left(a + \dfrac 1 {4a}\right) + \left(a - \dfrac 1 {4a}\right)=2a$
Thankyou very much. That helped a lot!