How do you simplify the expression #(cottsec^2t-cott)/(sint tant+cost)#?

1 Answer
MathFact-orials.blogspot.com
Dec 1, 2016

#sint#

Explanation:

(I'm writing this bit - the intro - last. I've just worked the problem and I had no idea where it would go or how to simplify it, beyond the fact that I took it apart piece by piece and took substitution opportunities where I could. It's easy to look at something complicated and shrug and think "I can't do this", but if you can start by doing something - anything - you can start to make sense of it. This is not to say that the first few things you do will make the problem simple right away - it can get rather messy. But stick with it and it should resolve itself into something simpler!)

Let's start with the original:

#(cottsec^2t-cott)/(sint tant+cost)#

In the numerator, let's factor out #cott# to get to a #(sec^2t-1)# term and in the denominator let's look at multiplying the sin and tan so that we can work with the cos:

#(cott(sec^2t-1))/(sint (sint/cost)+cost)#

In the numerator, we can use the trig identity #tan^2t=sec^2t-1# and substitute in. In the denominator, I'm going to work on getting the terms to add:

#(cott(tan^2t))/((sin^2t/cost)+cost(cost/cost))#

#((1/tant)(tan^2t))/((sin^2t/cost)+(cos^2t/cost))#

#tant/((sin^2t+cos^2t)/cost)#

#(tantcost)/(sin^2t+cos^2t)#

We can substitute in the denominator, using the trig identity #sin^2t+cos^2t=1#

#((sint/cancelcost)cancelcost)/1#

#sint#

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