Ответ:

Упростить выражение. Применяем основные тригонометрические тождества :    [tex]\bf 1+tg^2a=\dfrac{1}{cos^2a}\ \ ,\ \ 1+ctg^2a=\dfrac{1}{sin^2a}[/tex]   .

[tex]\bf \Big(2+tg^2a+ctg^2a\Big)\cdot tg^2a=\Big((1+tg^2a)+(1+ctg^2a)\Big)\cdot tg^2a=\\\\\\=\Big(\dfrac{1}{cos^2a}+\dfrac{1}{sin^2a}\Big)\cdot tg^2a=\dfrac{sin^2a+cos^2a}{cos^2a\cdot sin^2a}\cdot \dfrac{sin^2a}{cos^2a}=\\\\\\=\dfrac{1}{sin^2a\cdot cos^2a}\cdot \dfrac{sin^2a}{cos^2a}=\dfrac{1}{cos^4a}[/tex]

Преобразуем основное тригонометрическое тождество:

[tex] \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1 \: \: | \div \cos {}^{2} ( \alpha ) \\ \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } + \frac{ \cos {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } = \frac{1}{ \cos {}^{2} ( \alpha ) } \\\displaystyle\bf \tan {}^{2} ( \alpha ) + 1 = \frac{1}{ \cos {}^{2} ( \alpha ) } [/tex]

[tex] \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1 \: \: | \div \sin {}^{2} ( \alpha ) \\ \frac{ \sin {}^{2} ( \alpha ) }{ \sin {}^{2} ( \alpha ) } + \frac{ \cos {}^{2} ( \alpha ) }{ \sin {}^{2} ( \alpha ) } = \frac{1}{ \sin {}^{2} ( \alpha ) } \\ \displaystyle\bf 1 + \cot {}^{2} ( \alpha ) = \frac{1}{ \sin {}^{2} ( \alpha ) } [/tex]

[tex](2 + \tan {}^{2} ( \alpha ) + \cot {}^{2} ( \alpha ) ) \times \tan {}^{2} (a) = \\ = ( \tan {}^{2} ( \alpha ) + 1 + 1 \cot {}^{2} ( \alpha ) ) \times \tan {}^{2} ( \alpha ) = \\ = ( \frac{1}{ \cos {}^{2} ( \alpha ) } + \frac{1}{ \sin {}^{2} ( \alpha ) } ) \times \tan {}^{2} ( \alpha ) = \\= \frac{ \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) \sin {}^{2} ( \alpha ) } \times \frac{ \sin {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) } = \\ = \frac{1}{ \cos {}^{2} ( \alpha ) } \times \frac{1}{ \cos {}^{2} ( \alpha ) } = \frac{1}{ \cos {}^{4} ( \alpha ) } [/tex]