find tan with given cos and sin?

we know that 1+(tan x)^2=1/((cosx)^2)
and cos x>0
and tan x=sinx/cosx

draw a right triangle, with one leg 7 and the other 25. angle A is opposite the 7, so tan A = 7/25. the hypotenuse is √(7² + 25²) = √674, so
sin A = 7/√674 and cos A = 25/√674

You have two side lengths: 7, and 25.
Use the Pythagorean formula to solve for the third side = √(7² + 25²).

sin A = 7/(third side length)

cos A = 25/(third side length)

NOTE: the previous ANSWERS that say SINA = 7 and COSA = 25 ARE TOTALLY wrong.
Reason is that SIN and COS CAN NEVER be greater than ONE or less than NEGATIVE one. thats why their graphs look like, otherwise, test random values in calculator. BUT TAN CAN be greater than 1 or less than -1

My answer .......................................
given 0<A<90
you know this is a right angle trigonometry problem

tan A = opposite/adjancet = 7/25
thus you can say opposite is 7 and adjacent is 25 (remember they arent exactly these values, becasue tan sine and cosine are only ratios)

and hypotenuse = sqrt (7^2+25^2)

sinA = opp/hyp = 7/sqrt (7^2+25^2)
cosA = adj/hyp = 25/sqrt (7^2+25^2)

ALL YOU NEED TO REMEMBER for your test tomolo is this

remember TAN is O/A
and COS is A/H

Cos starts with C which is much closer to A than is S for sin to O thus you automatically know SIN is O/H

tanA= sinA/cosA

SinA=7

cosA=25

Solve the abv equalities to get the ans.

opposite = 7, adjacent = 25

hypotenuse = sqrt(7^2 + 25^2) = 25.96

sin = 7/25.96, cos = 25/25.96

TAN = SIN/COS Therefore SINA=7 while COSA=25.