Guys, just wanna know how all of you tackled this chapter as it contains too much formulas with conditions.
@Chirag_Hegde, @arush_kumar_singh, @pratyaksh_tyagi, @Viraam_Rao, @Lavesh_Gupta
Relate it with trigonometry.
Tan (a+b)= tanA+tanB/ 1-tanAtanB.
Similarly, tan"A + tan"B = tan" (A+B/1-AB)
But remember the conditions, they are very important.
What all formulas are troubling you?
For conditions for finding some of tan"a+tan"b I have a method rather than remembering conditions
Look what we do is for tan"a +tan"b
We get it by tan"(a+b)/1-ab
In this we get ans bw -π/2 to π/2
Where tan"a +tan"b can be in be -π to π
Here you need to look by approximate value like tan"1.2+tan"1.73 here if you apply above formula you will get ans in negative which isn't appropriate as both tan" are in bw 0 to π/2
Here we can solve it by approximation
Tan"1.2 is some angle greater than 45° and tan"1.73 is 60° so some will be greater than 90° tan"(a+b)/1-ab won't give you that so modify it i.e. if you are getting -theta then right is as π+(-∆) as tan (π+anglr) is same as tan(angle)
π+theta(it will come out to be negative) will be ans for tan"a +tan"b if both angle sum is greater than π/2
If it is in bw -π/2 to π/2 then and theta
If sum is < -π/2 then again π+theta
Similar approach for other problems of sum of angles
We were told not to learn the formulas, instead we were told to derive the formulas at home atleast once everyday for one week. The conditions and formulas got fit in my brain easily by 'repetition'.
@Mahesh_Kadam and @Manan_Upadhyay 's suggestions are better.
Yeah, My sir told the same thing to practice at home and remember the similarities in their proofs. So, you can easily derive them.
Practice and you'll realise that, its not at all difficult.