What is the zero power + Tan 60 degree of 0.25 times (- cos60 degrees) minus the zero power of (3-1 under the root sign)

What is the zero power + Tan 60 degree of 0.25 times (- cos60 degrees) minus the zero power of (3-1 under the root sign)

0.25 times (- cos60 degrees) minus the zero power of (3-1 under the root) + Tan 60 degrees
=0.25 / (cos60) - 1 + tan60
=0.25 / 0.25-1 + root 3
=Radical 3

sin² 30°-cos45° ×tan60°

sin² 30°-cos45° ×tan60°
=1/4-(√2/2)x(√3)
=1/4-√6/2

sin²25°+2sin60°+tan45°-tan60°+cos²25°=

solution
simple form
=(sin²25+cos²25)+2sin60+tan45-tan60
=1+2×(√3/2)+1-√3
=1+√3+1-√3
=2

cos²45º+tan30º.sin60º＝

cos²45º+tan30º.sin60º
=（√2/2）²＋√3×√3/2
=1/2+3/2
=2

It was proved that tan95 ° - tan35 ° - Radix 3 = 3tan95 ° tan35 °

tan(95-35)=[tan95-tan35]/(1+tan95*tan35)
Tan60 = radical 3 = (tan95-tan35) / (1 + tan95 * tan35)
Root number 3 + root number 3tan95 * tan35 = tan95-tan35
Tan95 ° - tan35 ° - root number 3 = root number 3tan95 ° tan35 °

tan95°−tan35°− 3tan95°tan35°=______ .

tan95°−tan35°−
3​tan95°tan35°
=tan(95°−35°)(1+tan95°tan35°)−
3​tan95°tan35°
=tan(60°)(1+tan95°tan35°)−
3​tan95°tan35°
=
3+
3tan95°tan35°−
3​tan95°tan35°
=
Three
So the answer is:
3．

Process and answer of tan25 + tan35 + tan25tan35

The title should be tan25 ° + tan35 ° + √ 3 tan25 ° · tan35 °. Tan25 ° + tan35 ° = Tan (25 ° + 35 °) · (1-tan25 ° tan35 °) + √ 3 tan25 ° · tan35 ° = √ 3 (1 - tan25 ° · tan35 °) + √ 3 tan25 ° · tan35 ° = √ 3

Reduction (tan5 π / 4 + tan5 π / 12) / (1-tan5 π / 12) Pay attention to the denominator. Be careful!

tan5π/4=tanπ/4
So we use the sum angle formula (tan45 + tan5 π / 12) / (1-tan45 * tan5 π / 12)
So it's equal to - radical 3
tan45=1 tan5π/12*1=tan5π/12*tan45
What's wrong

Reduction (tan5 π / 4 + tan5 π / 12) / (1-tan5 π / 12) (tan 5π/4 + tan 5π/12)/(1-tan 5π/12) =(tan π/4 + tan π/6+π/4 )/(1-tan π/4*tan π/6+π/4 ) =tan(π/4 + π/6+π/4 ) =tan(π/2+π/6) What's wrong with my calculation? Isn't Tan π / 2 meaningless?

......
=Tan (π / 2 + π / 6) you can do it here, all right
Tan (π / 2 + π / 6) ≠ Tan (π / 2), and Tan (π / 2 + π / 6) can not be expanded by trigonometric function formula of sum of two angles, but can be solved by the induction formula of trigonometric function of arbitrary angle
tan(π/2+π/6)=-cotπ/6=-√3.

The value of (1-tan5 π / 12tan π / 4) / (tan5 π / 12 + Tan π / 4)

The original formula = 1 / [(tan5 π / 12 + Tan π / 4) / (1-tan5 π / 12tan π / 4)]
=1/tan(5π/12+π/4)
=1/tan(2π/3)
=1/(-√3)
=-√3/3