1 +tan30÷1-tan30=1+sin60÷1-sin30
can anyone help me in proving this?​

Very quickly: since tan(30°) = 1/√3,

(1 + tan(30°)) / (1 - tan(30°)) = (1 + 1/√3) / (1 - 1/√3)

… = (√3 + 1) / (√3 - 1)

… = (√3 + 1)² / ((√3)² - 1²)

… = ((√3)² + 2√3 + 1²) / (3 - 1)

… = (3 + 2√3 + 1) / 2

… = (4 + 2√3) / 2

… = 2 + √3

On the other side, we have sin(30°) = 1/2 and sin(60°) = √3/2, so

(1 + sin(60°)) / (1 - sin(30°)) = (1 + √3/2) / (1 - 1/2)

… = (1 + √3/2) / (1/2)

… = 2 (1 + √3/2)

… = 2 + √3

and we're done, both sides are equal.

We can also prove this algebraically:

Force the denominator on the left side into a difference of squares:

(1 + tan(30°)) / (1 - tan(30°)) × (1 + tan(30°)) / (1 + tan(30°))

… = (1 + tan(30°))² / (1 - tan²(30°))

Expand the numerator:

… = (1 + 2 tan(30°) + tan²(30°)) / (1 - tan²(30°))

Recall that 1 + tan²(x) = sec²(x) :

… = (sec²(30°) + 2 tan(30°)) / (1 - tan²(30°))

Since sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x), multiply through every term by cos²(30°) :

… = (sec²(30°) + 2 tan(30°)) / (1 - tan²(30°)) × cos²(30°) / cos²(30°)

… = (1 + 2 sin(30°) cos(30°)) / (cos²(30°) - sin²(30°))

Recall that sin(2x) = 2 sin(x) cos(x) and cos(2x) = cos²(x) - sin²(x) :

… = (1 + sin(2×30°)) / cos(2×30°)

… = (1 + sin(60°)) / cos(60°)

Finally, recall that cos(90° - x) = sin(x), so cos(60°) = sin(30°) = 1/2, and we can write 1/2 = 1 - 1/2, so

… = (1 + sin(60°)) / (1 - sin(30°))

as required.

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

b

step-by-step explanation: