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ALGEBRA is great fun

Algebra is great fun - you get to solve puzzles!
With computer games you play by running, jumping or finding secret things. Well, with Algebra you play with letters, numbers and symbols, and you also get to find secret things!
And once you learn some of the "tricks", it becomes a fun challenge to work out how to use your skills in solving each "puzzle".equation
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how is tan (A+B) = [tan A + tan B]/[1 - tan A tan B]?

How is tan (A+B) = [tan A + tan B]/ [1 - tan A tan B]? Method 1: Let A = 30 deg and B = 45 deg. LHS = tan (30+45) = tan 75 = 3.732050808 RHS = [tan A + tan B]/[1 - tan A tan B] = [tan 30 + tan 45]/[1 - tan 30 tan 45] = [0.577350269 + 1]/[1 - 0.577350269*1] = 1.577350269/0.42264973 = 3.732050808 = LHS Proved. Method 2: tan (A+B) = [tan A + tan B]/[1 - tan A tan B] RHS = [tan A + tan B]/[1 - tan A tan B] =[(sin A/cos A) + (sin B/cos B)]/[1-(sin A/cos A)(sin B/cos B) = [sin A cos B + cos A sin B]/[cos A cos B][1 - sin A sin B/(cos A cos B)] = sin (A+B)/{[cos A cos B][cos A cos B - sin A sin B]/(cos A cos B)} = sin (A+B)/[cos A cos B - sin A sin B = sin (A+B)/cos (A+B) = tan (A+B) = LHS. Proved. Thanks. source:Quara
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What are some mind-blowing facts about mathematics?

Prime Generating Polynomials : The polynomial,  n 2 + n + 41 n 2 + n + 41  can be used to produce 40 primes for consecutive integer values 0≤n≤39.  This property was discovered by Euler . Similarly, the incredible formula,  n 2 − 79 n + 1601 n 2 − 79 n + 1601  was discovered, which produces 80 primes for the consecutive values 0≤n≤79! Kaprekar’s Constant : 6174 is known as Kaprekar’s Constant , after the Indian Mathematician D.R. Kaprekar. Take any four-digit number, using at least two different digits (Leading zeros are allowed.) Arrange the digits in descending and then in ascending order, to get two four-digit numbers, adding leading zeroes if necessary. Subtract the smaller number from the larger number and go back to step 2. The above process will always reach the fixed number, 6174, taking at most 7 iterations. Try it yourself! ( Note that after the first subtraction or the subsequent subtractions, the result obtained is always a multi...
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What are the last 3 digit of 2^2017?

I will provide two methods for this. Method 1  (Easy way) : USE A CALCULATOR. You will get the answer in a matter of seconds like I got. Clearly, the answer is 072. Method 2  (Slight harder way) : Here, I want to find the answer without using the calculator. Let's try it. Note :- I will be making use of  Congruence Modulo  and  Euler's Theorem , so these are the prerequisites. Another way to put the question is  “Find the remainder when  2 2017 2 2017  is divided by 1000”. First of all we factorise 1000 as: 1000 = 2 3 × 5 3 1000 = 2 3 × 5 3 Next, we find the remainder by  2 3 2 3  and  5 3 5 3 seperately. It's obvious that  2 3 2 3  divides  2 2017 2 2017 . Hence, 2 2017 ≡ 0 ( m o d 8 ) 2 2017 ≡ 0 ( m o d 8 ) Now to find the remainder by 125 (or 5^3), we use Euler's Theorem. Euler's theorem is applicable in this case since  g c d ( 2 , 125 ) = 1 g c d ( 2 , 125 ) = 1 . Φ ( 125 ) = ...
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