Good work on both. See below for the second.
The first looks very good. As long as you don't believe that cos(ab) = cos(a) cos(b), and as long as it doesn't confuse the question (as it would, for example, if cos(ab) was in fact part of the stated problem) you can get away with that shorthand.
Question: Confirm or refute,
Sec (a-b)= (Sec(a) Sec(b)) / (1 +Tan a Tan b)
Answer:
Right side = (Sec(a) Sec(b)) / (1 +Tan a Tan b)
Therefore:
(1/Cos a * 1/cos b) / (1 +(sin a / cos a) ( sin b/cos b ) )
(1/ cos ab ) / ( 1+ sin ab / cos ab )
(1/ cos ab ) / ( (Cos ab / (cos ab) + ( sin ab/cos ab ) )
(1/ cos ab ) / ( (cos ab + sin ab ) / ( cos ab ) )
(1/ cos ab ) * ( ( cos ab ) / (cos ab + sin ab) )
1 / (cos ab + sin ab)
Since :
1 / (cos ab + sin ab) = 1 / ( cos a cos b + sin a sin b )
AND
(Sec(a) Sec(b)) / (1 +Tan a Tan b) = 1 / ( cos a cos b + sin a sin b )
Therefore:
Sec (a-b) = (Sec(a) Sec(b)) / (1 +Tan a Tan b)
Because,
Sec (a-b) = 1 / ( cos a cos b + sin a sin b )
Sir,
For the following question what would be the proper way to refute this question.
Question:
(cos(a+b))/(Cos(a) Sin(b)) = 1-tan(a+b)
Ans:
(cos(a+b))= Cos(a) Cos (b) – Sin(a) Sin(b)
Therefore:
[Cos(a) Cos (b) – Sin(a) Sin(b)] / (Cos(a) Sin(b))
So,
(Cos (b)) / (Sin(b)) – (sin(a)) / (Cos(a))
1/tan(b) –Tan (a)
= (1-Tan(ab)) / Tan (b)
accepting that by tan(ab) you really mean tan(a) * tan(b), as seems clear below ...
Therefore:
1/Tan(b) – (Tan(a)/Tan(b) ) ( Tan(b) /Tan(b) )
Equals …. (1-Tan (a) ) / Tan (b)
If a= 0 then we are left with (1/ Tan (b)) and the identity must be true for all values…
Therefore, I would refute
(cos(a+b))/(Cos(a) Sin(b)) = 1-tan(a+b)
because the left hand side equals .. (1-Tan(ab)) / Tan (b)
What’s your thoughts
I can't find a more elegant way to prove this. Your last statement, however, doesn't quite nail it down. How about the following?
The left-hand side can indeed be reduced to
1/tan(b) –Tan (a)
and the right-hand side is 1 - tan(a+b) so the identity becomes
1/tan(b) –Tan (a) = 1 - tan(a+b) .
If a = 0 this becomes just
1 / tan(b) = 1 - tan(b).
In general the reciprocal of a number is not equal to 1 - that number. More specifically we could rearrange this to read
1 = tan(b) - tan^2(b) so that
tan^2(b) - tan(b) + 1 = 0. The quadratic formula tells us that this can be true for at most two values of tan(b) (in this case the discriminant is negative so there are in fact no such values), so clearly it cannot be so for all values of b.