<s>[INST] 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.
 [/INST] Too long for a comment:
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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
$$\csc\theta+\cot\theta=\frac{13+5}{12}=\frac{3}{2}.$$
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