Finding tanα if tan(α+β)=7 and tan(α-β)=5

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[vardhanr]

Another method (the way I did it):
α + β = atan(7)
α - β = atan(5)
Add both of these equations to get:
2α = atan(7) + atan(5)
==> tan(2α) = tan(atan(7) + atan(5))
==> 2tan(α) / (1 - tan ^ 2 (α)) = (7 + 5) / (1 - 7 * 5)
==> 17tan(α) = 3tan ^ 2 (α) - 3
==> 3tan ^ 2 (α) - 17tan(α) - 3 = 0
==> tan(α) = (17 +/- 5sqrt(13)) / 6

[jasoncetron]

Small error at the end, which is something I do way too often with my students: The 17 in the denominator "turned into" a 7. This usually happens at the very end of the period when we're rushing to fit in the entire solution to a problem. Inevitably, with seconds left, a compassionate student will interrupt the frantic search for the mathematical error and say "Wait! You just forgot to write the 1". :-)

[MushookieMan]

119 is what I call a "surprise composite number", because it is an odd number less than 144, that has a prime factor greater than 12, and doesn't end in 5. Long story short, we only memorize times tables up to 12. 51 and 57 are also surprising to me.

[SanjuSingh-pprs]

Easiest Method:

First use the formulas of tan(a + b) and tan(a - b)
-> (tan(a) + tan(b))/(1 - tan(a)*tan(b)) = 7 --(eq-1)
-> (tan(a) - tan(b))/(1 + tan(a)*tan(b)) = 5 --(eq-2)

Then multiply the denominators of both equations with 7 and 5 respectively, we get

tan(a) + tan(b) = 7 - 7*tan(a)*tan(b) --(eq-3)
tan(a) - tan(b) = 5 + 5*tan(a)*tan(b) --(eq-4)

Then add (eq-3) and (eq-4), we get

=> 2*tan(a) = 12 - 2*tan(a)*tan(b)
=> 2*tan(a)(1 + tan(b)) = 12
=> tan(a)(1 + tan(b)) = 6 --(eq-5)

from (eq-3)
=> tan(a)(1 + 7*tan(b)) = 7 --(eq-6)

Dividing (eq-6) & (eq-5), we get

=> (1 + 7*tan(b))/(1 + tan(b)) = 7/6

Cross multiplying the equation, we get

=> 6 + 42*tan(b) = 7 + 7*tan(b)
=> 35*tan(b) = 1
=> tan(b) = 1/35

Put the value of tan(b) in (eq-5), we get

=> tan(a)(1 + (1/35)) = 6
=> tan(a)(36/35) = 6
=> tan(a) = 35/6

Hence, the value of tan(a) = 35/6

This is the easiest way to solve this question without any mess and in less than 1 minute.

Do comment if you found this method helpful😄😄😄

[agni]

Hey just found your channel...love these nice little(sometimes not so much) math problems..great work!

[udic]

11:25 your 17 turned by mistake to 7...

[johngreen]

The coefficient linear term on the second quadratic should be -17 not -7, you dropped the tens digit of 1. Its a transcription error.

[sohelzibara]

here is another approach:
Find tan b in two different ways as shown down
1) tan b = tan (b + a - a) = (7 - tan a)/(1 + 7 tan a)
Similarly
2) tan b = tan (b - a + a) = (-5 +tan a)/(1 + 5 tan a)
equating yields
3 tan²a - 17 tan a- 3 = 0
and the rest follows.

[BlackwoodCompany]

It was simpler than I thought with linear equations due to the inverse tangent being a constant no matter in radians or degrees (I use degrees most of the time)

It was doing a + b = tan -1 (7), and the a - b = tan -1 (5). This is when adding them on both sides cancels out b leaving x = (tan -1 (7) + tan -1 (5))/2. Then it is plugging in this one, and taking the tangent of that answer.

[king_haku]

loved the second solution! very simple and elegant.

[richardryan]

Very nice! By the way, a 12.6 minute video isn't too long. We have excellent attention spans. 8-)

[michaelpurtell]

Did it by the first method which I considered ugly . Sensed that you must have a “better” approach, and you awarded me with the second method, which I consider elegant

[vimleshmaheshwari]

Divide both tan(A+B)/tan(A-B)=7/5

Now apply componendo dividendo
We can get the answer

[wesleysuen]

Note that the two possible values for tan(alpha) are in fact negative reciprocals of each other, corresponding to the principal value of alpha (about 80.27998 degrees) plus any integral multiples of 90 degrees.

[unonovezero]

325 ending with 25, it's a multiple of 25, you can't miss it 😊

[RCSmiths]

Trig puzzles and problems are always interesting. Cheers!

[oguzhanozdogan]

If angles are written in triangle ABC, from the bisector theorem √26/√50=a^2/(a-2)^2
a=(-13+-(√325)/6
tan(alpha)=a+5=(17+-(√325)/6)

[kmlhll]

thank you very much for your very simple and fast methods, I kindly ask you to include in your videos examples of integral calculations using limit development. thank you.

[ananthiyengar]

Your way of teaching is excellent, and creat interest in maths, thank you v much

[ghanshyamsharma]

In second method 17 is written as 7 (-6/17 is written as -6/7)