How do you verify the identity (tan^(2)x+2tanx+1)/(sinxtanx+sinx)?
1 Answer
This expression is equal to
Explanation:
Let's look at what we have to start with, and how we can simplify it:
First, remember the formula for the square of a binomial:
#(a+b)^2 = a^2 + 2ab + b^2# This looks A LOT like the numerator of our expression!
#(tanx + 1)^2 = tan^2x+2(tanx)(1) + 1^2# In fact, the top part IS the square of
#(tanx+1)# , as verified by this rule. So we can simplify the numerator as such.
Next, notice that both terms on the bottom have
#sinx# in them. So, we can factor out a#sinx# from both of them:
#sinxtanx+sinx = sinx(tanx)+sinx(1) = sinx(tanx+1)# Therefore, we can factor the denominator into
#sinx(tanx+1)#
Both the numerator and denominator have a factor of
#tanx+1# , so we can cancel them out!
#(tanx+1)^2/(sinx(tanx+1))#
#((tanx+1)(tanx+1))/(sinx(tanx+1))#
#((tanx+1)cancel((tanx+1)))/(sinxcancel((tanx+1))#
Next, let's split the fraction into two, since the numerator has two terms.
We know that
#tanx = sinx/cosx# . Therefore:
#tanx/sinx = (sinx/cosx)/sinx = (cancel(sinx)/cosx)/cancel(sinx) = 1/cosx = secx# We also know that
#1/sinx = cscx# Therefore, we can simplify both of these fractions and get the simplest form of our expression:
Final Answer
