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.

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: