A) \[{{(2+{{e}^{2}})}^{\frac{3}{2}}}\]
B) \[{{\left( 2-{{e}^{2}} \right)}^{\frac{1}{2}}}\]
C) \[{{\left( 2+{{e}^{2}} \right)}^{\frac{1}{2}}}\]
D) \[{{(2-{{e}^{2}})}^{\frac{3}{2}}}\]
Correct Answer: D
Solution :
(d): \[ta{{n}^{2}}x=1-{{e}^{2}}\] \[\therefore secx+ta{{n}^{3}}x.cosecx\] \[=secx+ta{{n}^{2}}x.\,tanx.\text{ }cosecx\] \[=secx+ta{{n}^{2}}x.\frac{\sin x}{\cos x}.\frac{1}{\sin x}\] \[=secx+ta{{n}^{2}}x.\text{ }secx\] \[=secx.\left( 1+ta{{n}^{2}}x \right)\] \[={{\left( 1+{{\tan }^{2}}x \right)}^{\frac{1}{2}}}.\left( 1+{{\tan }^{2}}x \right)\] \[={{(1+{{\tan }^{2}}x)}^{\frac{3}{2}}}={{(1+1-{{e}^{2}})}^{\frac{3}{2}}}\] \[={{\left( 2-{{e}^{2}} \right)}^{\frac{3}{2}}}\]
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