#( 1+ cot(\theta )- csc(\theta ) )( 1- tan(\theta )+ sec(\theta ) )= 2# ?

1 Answer
Feb 17, 2018

Answer:

The solution to #Equation=2#
is
#theta=pi/2# (modulo #2 pi#).

HOWEVER, I REALLY THINK THERE IS A MISTAKE IN THE EQUATION AS THINGS WOULD CANCEL OUT NICELY IF THE EQUATION IS THE FOLLOWING:
#Equation=(1 + cot(theta) - csc(theta))(1+tan(theta)+sec(theta))#

(should be #+tan(theta)# instead of #-tan(theta)#).

Explanation:

Since the question is vague, I would just try to simplify the equation a bit first.
We start with:
#Equation=(1 + cot(theta) - csc(theta))(1-tan(theta)+sec(theta))#

and rewrite everything in terms of sine and cosine:
#=(1+cos(theta)/sin(theta) - 1/sin(theta)) (1-sin(theta)/cos(theta) + 1/cos(theta))#

Simplify a bit more:
#=(1+ (cos(theta) - 1)/(sin(theta))) (1+(1-sin(theta))/(cos(theta)))#

Opening up the parentheses:
#=1+(1-sin(theta))/(cos(theta))+(cos(theta) -1)/(sin(theta))+((cos(theta)-1)(1-sin(theta)))/(sin(theta)cos(theta))#

And putting everything under the same denominator gives:
#=(sin(theta)cos(theta)+sin(theta)(1-sin(theta))+cos(theta)(cos(theta)-1)+(cos(theta)-1)(1-sin(theta)))/(sin(theta) cos(theta))#

Now, we look just at the numerator:
#=sin(theta)cos(theta)+sin(theta)-sin^2(theta)+cos^2(theta)-cos(theta)+cos(theta)-sin(theta)cos(theta)-1+sin(theta)#
#=2sin(theta) -sin^2(theta) + cos^2(theta)-1#
#=2sin(theta)-sin^2(theta)+cos^2(theta)-sin^2(theta)-cos^2(theta)#
#=2sin(theta)(1-sin(theta))#

Now, back to the equation:
#Equation=(2sin(theta)(1-sin(theta)))/(sin(theta) cos(theta))#
So,
#Equation=2((1-sin(theta)))/(cos(theta))#

Now, let's solve for theta when #Equation =2#
This happens when #(1-sin(theta))/(cos(theta))=1#
i.e.
#1-sin(theta)=cos(theta)#
i.e.
#sin(theta)+cos(theta)=1#
This only happens when either #sin(theta)=1# and #cos(theta)=0#
when #theta=pi/2# (modulo #2pi#)
or when #sin(theta)=0# and #cos(theta)=1# when #theta=0# (modulo #2pi#).
Since #Equation =2((1-sin(theta)))/(cos(theta))#, #cos(theta)# cannot be 0 and
thus the solution to #Equation=2#
is
#theta=pi/2# (modulo #2 pi#).

HOWEVER, I REALLY THINK THERE IS A MISTAKE IN THE EQUATION AS THINGS WOULD CANCEL OUT NICELY IF THE EQUATION IS THE FOLLOWING:
#Equation=(1 + cot(theta) - csc(theta))(1+tan(theta)+sec(theta))#
(#+tan(theta)# instead of #-tan(theta)#).
Then, the equation is indeed 2 for all #theta#.