Question

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(tan theta + sec theta)^2

Answer 1

\[ \left( \tan \theta + \sec \theta \right)^2\]

Answer 2

Step-by-step, please check me, but I think this is good:
\[(\sec(\theta)+\tan(\theta))^2\]
\[(\sec(\theta)+\tan(\theta)) (\sec(\theta)+\tan(\theta))\]
\[\sec(\theta) (\sec(\theta)+\tan(\theta))+\tan(\theta) (\sec(\theta)+\tan(\theta))\]
\[\sec(\theta) (\sec(\theta)+\tan(\theta))+(\sec(\theta)+\tan(\theta)) \tan(\theta)\]
\[\sec(\theta) \sec(\theta)+\sec(\theta) \tan(\theta)+(\sec(\theta)+\tan(\theta)) \tan(\theta)\]
\[\sec(\theta) \sec(\theta)+\sec(\theta) \tan(\theta)+\sec(\theta) \tan(\theta)+\tan(\theta) \tan(\theta)\]
\[\sec(\theta) \sec(\theta)+2 \sec(\theta) \tan(\theta)+\tan(\theta) \tan(\theta)\]
\[\sec^{1+1}(\theta)+2 \sec(\theta) \tan(\theta)+\tan^{1+1}(\theta)\]
\[\sec^2(\theta)+2 \sec(\theta) \tan(\theta)+\tan^2(\theta)\]
If you have questions as to what I did on a particular step please ask away!

Answer 3

yep, it's right, if you want too simplify more I think you can put it as ((sin(theta)+1)/cos(theta))^2

Answer 4

Thanks all xD

Answer 5

:-)

Answer 6

yu have any doubt???????????

Answer 7

Because I can make typos :D