The first line:

1. Using trigonometric function and differential product formula, sin15cos5-sin20 =1/2 (sin20+sin10)-sin20 =1/2 (-sin20

2. Similarly, cos15cos5-cos20 = sin5sin15.

3. So the original formula = 1/tan 15 =-(2+ root number 3).

The second way:

Since everyone knows better, I won't say more. Just a reminder, the key is triangle substitution.

The third way:

1. Note that A+B+C = 180, C = 90, and G therefore A+B = 90, so COSA COSB = COSA COS (90-A) = COSA SINA =1/2sin2A.

2. Pay attention to the value of sine function. A is at (0,90), so cosAcosB is at (0, 1/2).

The fourth way: (Vieta's theorem should be found: tanA+tanB=-5/3, tanAtanB=-7/3. )

( 1)

sin(A+B)/COS(A-B)=(Sina cosb+cosa sinb)/(cosa cosb+Sina sinb)=(tanA+tanB)/( 1+tanA tanB)= 5/4 .

(2)

1. Note that the square of tanA+the square of Tanb = (Tana +tanB)-2 the square of Tana Tanb =67/9.

2. The square of COS (a+b) = the square of COS (a+b)+the square of +cosA) (the square of SINA+the square of +cosB) = (the square of 65438+0+(tanatanb)-2 tanatanb)/(tanatanb). (Note: The second equal sign is obtained by dividing the numerator denominator by the square of (cosAcosB) at the same time. )

A little trouble, clear thinking.

What do you think?