Given $\sec \theta + \tan \theta = 5$ , Find $\csc \theta + \cot \theta $. The question is to find the value of $ \csc  \theta + \cot \theta $ if $\sec  \theta + \tan \theta = 5$ .
Here is what I did :
$\sec  \theta + \tan \theta = 5$
$\sec  \theta = 5 -  \tan \theta $
Squaring both sides , 
 $$\sec^2  \theta = 25  + \tan^2 \theta -10\tan \theta$$
Substituting $1+\tan^2 \theta$ for $\sec^2 \theta$ ,
$$1+\tan^2 \theta = 25 + \tan^2 \theta -10\tan \theta$$
Thus , $$\tan \theta=24/10$$
So , $\cot \theta = 10/24 $ and $\csc \theta=26/24$ 
Thus $ \csc  \theta + \cot \theta =3/2$ .
But I checked the answer sheet and the answer is not 3/2 but $(3+\sqrt5 )/2$ .
Where have I went wrong ? Please help.
 A: As noted in the comments, you are correct, and the answer key is wrong here.
In cases like these, it's sometimes helpful to check to make sure that you didn't make a mistake in reading the question.
A: Here is a simpler solution to this problem:
$$\left(\sec(\theta)+\tan(\theta) \right)\left(\sec(\theta)-\tan(\theta) \right)=\sec^2(\theta)-\tan^2(\theta)=1$$
Since $\sec(\theta)+\tan(\theta)=5$ you get $\sec(\theta)-\tan(\theta)=\frac{1}{5}$.
Adding and subtracting these two relations you get
$$2\sec(\theta)=5+\frac15=\frac{26}{5} \,;\, 2\tan(\theta)=5-\frac15=\frac{24}{5}$$
Thus $\tan(\theta)=\frac{24}{10}$ and 
$$\sin(\theta)=\frac{\tan(\theta)}{\sec(\theta)}=\frac{24}{26} \,.$$
A: The problem is that the methods you have give all ok so long it works but nothing if it fails.
When in doubt it is better to check with nearby approximate solutions learning to guide/trust yourself. 
Say even if you took very approximately 60 degrees as solution and calculated $  sec  (\theta) + tan(\theta) $ it will be nearer to what you got than the way off value   provided by key!
A: Too long for a comment:
...
THIS IS HOW I THINK THE ANSWER SHEET WRITER DID THE MISTAKE:
Normally the equation is
$$\sin\theta+1=5\cos\theta.$$
Squaring they  "mistakenly" got $\sin^2\theta+2\sin\theta+1=5\cos^2\theta\implies$ $\sin^2\theta+2\sin\theta+1=5-5\sin^2\theta\implies$ $3\sin^2\theta+\sin\theta-2=0\implies$ $(3\sin\theta-2)(\sin\theta+1)=0$.
Thus, $\sin\theta=\frac{2}{3}$ is a solution. Finally, $\csc\theta=\frac{3}{2}$, $\cot\theta=\frac{\sqrt{5}}{2}$ and $\csc\theta+\cot\theta=\frac{3+\sqrt{5}}{2}$.
CORRECT SOLUTION:
But in fact squaring we get $(13\sin\theta-12)(\sin\theta+1)=0$.
Thus, $\sin\theta=\frac{12}{13}$ is a solution. Finally, $\cot\theta=\frac{5}{12}$, $\csc\theta=\frac{13}{12}$ and
$$\csc\theta+\cot\theta=\frac{13+5}{12}=\frac{3}{2}.$$
