Question

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Answer 1

12.2

1.

${ \sin }^{2} (2 \alpha ) + { \cos}^{2} (2 \alpha ) + \frac{1}{ {tg}^{2} (5 \alpha )} = \\ = 1 + \frac{1}{ {tg}^{2}(5 \alpha ) } = 1 + {ctg}^{2} (5 \alpha ) = \\ = \frac{1}{ { \sin }^{2}(5 \alpha ) }$

2.

$(1 + \sin( \frac{x}{2} ) )(1 - \sin( \frac{x}{2} ) ) = \\ = 1 - { \sin}^{2} ( \frac{x}{2} ) = { \cos }^{2} ( \frac{x}{2} )$

3.

$(tg \alpha \cos( \alpha )) ^{2} + {( \frac{ \sin( \alpha ) }{tg \alpha } )}^{2} = \\ = ( \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \times \cos( \alpha )) ^{2} + {( \sin( \alpha ) \times \frac{ \cos( \alpha ) }{ \sin( \alpha ) }) }^{2} = \\ = \sin ^{2} ( \alpha ) + { \cos }^{2} ( \alpha ) = 1$

4.

$\frac{ { \sin }^{2}( \alpha ) {tg}^{2}( \alpha ) }{ {tg}^{2} ( \alpha ) + { \cos }^{2} ( \alpha ) - 1} = \frac{ { \sin }^{2} (\alpha) {tg}^{2}( \alpha )}{ {tg}^{2} (\alpha) - { \sin }^{2}( \alpha) } = \\ = { \sin }^{2} ( \alpha ) {tg}^{2} ( \alpha ) \times \frac{1}{ \frac{ { \sin }^{2} ( \alpha )}{ { \cos }^{2} (\alpha )} - { \sin }^{2} (\alpha ) } = \\ = { \sin }^{2} (\alpha ) {tg}^{2} ( \alpha ) \times \frac{ { \cos }^{2}( \alpha ) }{ { \sin}^{2}( \alpha ) - { \sin }^{2} ( \alpha) { \cos}^{2}( \alpha ) } = \\ = { \sin }^{4} (\alpha ) \times \frac{1}{ { \sin }^{2}( \alpha )(1 - { \cos }^{2}( \alpha )) } = \\ = \frac{ { \sin}^{2} (\alpha ) }{ { \sin }^{2} (\alpha ) } = 1$

4.