# 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.

• You went wrong trusting the answer sheet. Your answer is correct. Aug 10, 2013 at 17:49
• Indeed; if you happened to see it, please disregard my previous erroneous comment. Everything looks good here. Aug 10, 2013 at 17:50
• Is that really the case ? I'm surprised because the answer sheet is hardly wrong . Is there any online resource I can use to validate answers to questions like this ? Aug 10, 2013 at 17:53
• Or this followe by this (42 is arbitrary). Aug 10, 2013 at 18:05
• Don't delete the question---it's fine! However, 10/24 looks better as 5/12, etc. Aug 10, 2013 at 18:09

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

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.

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!

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 $$\csc\theta+\cot\theta=\frac{13+5}{12}=\frac{3}{2}.$$