<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] 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 
$$\sin(\theta)=\frac{\tan(\theta)}{\sec(\theta)}=\frac{24}{26} \,.$$
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