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(1+tanx+sec x)2+(1−tanx+secx)2 secx(1+secx)\frac{\left(1+tanx+sec\:x\right)^2+\left(1-tanx+secx\right)^2}{\:secx\left(1+secx\right)}

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Final answer to the exercise

1+2tan⁡(x)cos⁡(x)2+(1−tan⁡(x)+sec⁡(x))2cos⁡(x)2cos⁡(x)+1\frac{1+2\tan\left(x\right)\cos\left(x\right)^2+\left(1-\tan\left(x\right)+\sec\left(x\right)\right)^2\cos\left(x\right)^2}{\cos\left(x\right)+1}

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(1+tanx+sec x)2+(1−tanx+secx)2 secx(1+secx) 
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