\( \left\{ \begin{array} { l l } { 5.2 .2 } & { \frac { \cos ( - A ) \cdot \tan ( 180 ^ { \circ } - A ) \cdot \sin ( 180 ^ { \circ } - A ) } { \sin ( 360 ^ { \circ } - A ) \cos ( 90 ^ { \circ } - A ) } } \\ { 5.2 .3 } & { \cos ^ { 2 } ( 180 ^ { \circ } + x ) + \tan ( x - 180 ^ { \circ } ) \cdot \sin ( 720 ^ { \circ } - x ) \cdot \sin ( 90 - x } \\ { 5.2 .4 } & { \frac { \cos ( x - 180 ^ { \circ } ) \cdot \cos ( 90 ^ { \circ } - x ) } { \sin ( 90 ^ { \circ } + x ) \cdot \sin ( - x - 180 ) } } \\ { 5.2 .5 } & { \frac { \cos ( \alpha - 90 ^ { \circ } ) \cdot \tan ( - \alpha ) } { \sin ( - \alpha ) \cdot \tan ( 720 ^ { \circ } - \alpha ) } } \\ { 5.2 .6 } & { \frac { \sin ( \beta - 180 ^ { \circ } ) \cdot \tan ( - \beta - 180 ^ { \circ } ) \cdot \cos ( 180 ^ { \circ } + \beta ) } { \cos ( - \beta ) \cdot \sin ( 360 ^ { \circ } + \beta ) } } \end{array} \right. \)

We wish to simplify each of the following expressions using standard trigonometric identities. In what follows we assume all denominators are nonzero. ────────────────────────────── Problem 5.2.2 Expression:   [ cos(–A) · tan(180° – A) · sin(180° – A) ] / [ sin(360° – A) · cos(90° – A) ] Step 1. Rewrite each function:  • cos(–A) = cos A        (cosine is even)  • tan(180° – A) = –tan A     (tangent identity: tan(π – θ) = –tan θ)  • sin(180° – A) = sin A      (sine identity)  • sin(360° – A) = –sin A      (since sin(360° – A) = sin(–A) = –sin A)  • cos(90° – A) = sin A       (co‐function identity) Step 2. Substitute into the expression:  Numerator = (cos A) · (–tan A) · (sin A) = – cos A · tan A · sin A  Denom = (– sin A) · (sin A) = – sin² A Step 3. Simplify:   (– cos A · tan A · sin A) / (– sin² A)  Cancel the minus signs:   = (cos A · tan A · sin A) / (sin² A) Now, note that tan A = sin A / cos A. So,   = [cos A · (sin A/cos A) · sin A] / (sin² A)   = (sin² A) / (sin² A) = 1 Answer for 5.2.2: 1 ────────────────────────────── Problem 5.2.3 Expression:   cos²(180° + x) + tan(x – 180°) · sin(720° – x) · sin(90° – x) Step 1. Simplify each term:  • cos(180° + x) = – cos x      so cos²(180° + x) = cos² x  • tan(x – 180°) = tan x       (since tan has period 180°)  • sin(720° – x): Notice 720° = 2·360°, so sin(720° – x) = sin(–x) = – sin x  • sin(90° – x) = cos x       (co‐function identity) Step 2. Substitute:  Second term = tan x · (– sin x) · cos x = –tan x · sin x · cos x Now use tan x = sin x / cos x:   = – (sin x / cos x) · sin x · cos x = – sin² x Step 3. Combine with the first term:   cos² x + (– sin² x) = cos² x – sin² x But recall the double‐angle formula:   cos² x – sin² x = cos 2x Answer for 5.2.3: cos 2x ────────────────────────────── Problem 5.2.4 Expression:   [ cos(x – 180°) · cos(90° – x) ] / [ sin(90° + x) · sin(–x – 180°) ] Step 1. Rewrite each function:  • cos(x – 180°) = – cos x     (cosine shifted by 180°)  • cos(90° – x) = sin x       (co‐function)  • sin(90° + x) = cos x       (co‐function: sin(90°+θ)=cosθ)  • For sin(–x – 180°): write it as sin[–(x + 180°)].   Since sin(–θ) = – sin θ, then    sin(–x – 180°) = – sin(x + 180°)  • But sin(x + 180°) = – sin x, so   sin(–x – 180°) = – (– sin x) = sin x Step 2. Substitute:  Numerator = (– cos x) · (sin x) = – cos x · sin x  Denom = (cos x) · (sin x) = cos x · sin x Step 3. Simplify:  = (– cos x · sin x) / (cos x · sin x) = –1 Answer for 5.2.4: –1 ────────────────────────────── Problem 5.2.5 Expression:   [ cos(α – 90°) · tan(–α) ] / [ sin(–α) · tan(720° – α) ] Step 1. Rewrite each function:  • cos(α – 90°) = sin α       (since cos(θ – 90°) = sin θ)  • tan(–α) = – tan α        (tangent is odd)  • sin(–α) = – sin α        (sine is odd)  • tan(720° – α): Note 720° = 2·360° so tan(720° – α) = tan(–α). And tan(–α) = – tan α. Step 2. Substitute:  Numerator = sin α · (– tan α) = – sin α · tan α  Denom = (– sin α) · (– tan α) = sin α · tan α Step 3. Simplify:  = (– sin α · tan α) / (sin α · tan α) = –1 Answer for 5.2.5: –1 ────────────────────────────── Problem 5.2.6 Expression:   [ sin(β – 180°) · tan(–β – 180°) · cos(180° + β) ] / [ cos(–β) · sin(360° + β) ] Step 1. Simplify each factor:  • sin(β – 180°) = – sin β    Reason: sin(β – 180°) = sin β cos180° – cos β sin180° = (sin β)(–1) – (cos β)(0) = – sin β.  • tan(–β – 180°): Write –β – 180° = –(β + 180°).   Since tan(–θ) = – tan θ, this becomes – tan(β + 180°).   And because tangent has period 180°, tan(β + 180°) = tan β.   Thus, tan(–β – 180°) = – tan β.  • cos(180° + β) = – cos β  • cos(–β) = cos β       (cosine is even)  • sin(360° + β) = sin β     (sine has period 360°) Step 2. Substitute into the numerator:  Numerator = (– sin β) · (– tan β) · (– cos β)  Multiply the signs: (–)·(–)·(–) = –   So numerator = – sin β · tan β · cos β.  Now, note tan β = sin β / cos β, so   = – sin β · (sin β/cos β) · cos β = – sin² β Step 3. Denominator:  Denom = cos β · sin β Step 4. Simplify:  = (– sin² β) / (cos β · sin β) = – (sin β / cos β) = – tan β Answer for 5.2.6: – tan β ────────────────────────────── Summary of Answers: 5.2.2 = 1   5.2.3 = cos 2x   5.2.4 = –1   5.2.5 = –1   5.2.6 = – tan β These are the simplified forms of the given expressions.