How to get 1 + Tan square α = 1 / cos square α = 1

How to get 1 + Tan square α = 1 / cos square α = 1

Sin squared α + cos squared α = 1, divide the two sides by (COSA) ^ 2
(sina)^2/(cosa)^2+1=1/(cosa)^2
That is: 1 + (Tana) ^ 2 = 1 / (COSA) ^ 2

Calculation: Tan 15 ° / (1-tan 2 15 °)

tan30=√3/3=tan(2*15)=2tan15/(1-tan²15)
So tan15 / (1-tan? 15) = √ 3 / 6

As shown in the figure, we know that ab = ad, angle ABC = angle ADC, indicating the reason that BC = CD

Should be missing condition: AC ⊥ BD
In △ ABC and △ ADC
AB=AD,∠ABC=∠ADC,∠BAC=∠DAC
So △ ABC ≌ △ ADC
Thus BC = DC

As shown in the figure, given that AD / / BC, angle ADC = angle ABC, then AB / / CD? Try to explain the reason If you know what a graph is, don't ask if you don't know As shown in the figure, it is known that C, P and D are on the same straight line, the angle BAP is complementary to the angle APD, the angle CPE is equal to the angle fab, is the angle e equal to the angle f? Try to explain why As shown in the figure, angle 1 and angle 2 complement each other, angle B = angle D, verification: AB / / CD

Ad / / BC, so angle ADC + angle DAC = 180, angle DCB + angle ABC = 180
Angle ADC = angle ABC, then angle ADC + angle DCB = 180
So AB / / CD

As shown in the figure, ① To use "SAS" to explain △ ABC ≌ △ ADC, if AB = ad, the conditions need to be added are______ ; ② To use "ASA" to describe △ ABC ≌ △ ADC, if ∠ ACB = ∠ ACD, the conditions need to be added are______ .

(1) In △ ABC and △ ADC, ab = ad ∠ BAC = ∠ dacac = AC, ≌△ ADC (SAS); (2) add condition ∠ BAC = ∠ DAC. ∵ in △ ABC and △ ADC, ∵ ACB = acdac = AC ∠ DAC, △ ABC ≌ △ ADC (ASA)

As shown in the figure, in △ ABC, ad and be are the heights on BC and AC respectively, and AD and be intersect at point F. verification: △ AEF ∽ ADC It is known that: as shown in the figure, in △ ABC, ad and be are the heights on the edge of BC and AC respectively, and AD and be intersect at point F. (1) verification: △ AEF ∽ ADC (2) Are there any triangles similar to △ AEF in the figure? Please write down Urgent The graph is △ ABC, and then the high be on the AC edge intersects AC with E. BC, and the high ad intersects BC with D

1. In △ AEF, the angle AEF = 90 ° and in △ ADC, the angle ADC = 90 ° and the angle EAF = angle CAD. Therefore, △ AEF ∽ ADC
2.:△AEF∽△BDF.:△BDF∽△BCE

As shown in the figure, D and E in △ ABC are points on edge BC and ab respectively, and ∠ 1 = ∠ 2 = ∠ 3. If the perimeter of △ ABC, △ EBD and △ ADC is m, M1 and M2 in turn, it is proved that: M1 + m2 m≤5 4．

Let BC = a, AC = B, ∵ 1 = ∠ 2 = ∠ 3,
∴△ABC∽△EBD∽△DAC，
∴DC
AC=AC
BC，
∴DC=b2
a，BD=BC-DC=a-b2
a=a2−b2
a，
∵m1
m=BD
BC=a2−b2
a2，m2
m=AC
BC=b
a，
∴m1+m2
m=a2−b2
a2+b
a=-（b
A-1
2）2+5
4≤5
4．

As shown in the figure, in triangle ABC, D and E are the points on side BC and ab respectively, and angle 1 = angle 2 = angle 3. The perimeter of triangle ABC, triangle EBD and triangle ADC is in turn m. N, P, prove that (n + P) / M is less than or equal to 5 / 4

Analysis: let BC = a, AC = B, from ∠ L = ∠ 2 = ∠ 3, we can get △ ABC ∽ △ EBD ∽ DAC

As shown in the figure, △ ABC, De is the vertical bisector of edge AB, ab = 6, BC = 8, AC = 5, then the circumference of △ ADC is () A. 14 B. 13 C. 11 D. 9

Because de bisects AB vertically, ad = BD, AC = 5, BC = 8,
So the circumference of △ ADC is AD + DC + AC = BD + DC + AC = BC + AC = 8 + 5 = 13 (CM)
Therefore, B

As shown in the figure, we know AB parallel CD, be bisection ∠ ABC, de bisection ∠ ADC, ∠ bad = 80 °, try to find: I can do it, I don't need it 2. If ∠ BCD = n °, try to find the degree of ∠ bed The picture is on the test paper and can't be sent out Solve

According to the polygon inner angle sum formula 180 ° (n-2)
So the sum of interior angles of quadrilateral ABCD and quadrilateral BCDE is 360 degrees
So ∠ ABC + ∠ CDA = 360-80-n = 280 ° - n °
So ∠ EBC + ∠ CDE = 140 ° - n ° / 2
So ∠ bed + BCD = 360 - (140-n / 2) = 220 ° + n ° / 2
∠BED=220°-n°/2