Asked by Ray
Simply:
cot(-x)-1/1-tan(-x)
Is it -cotx? The other answer choice is: cotx.
cot(-x)-1/1-tan(-x)
Is it -cotx? The other answer choice is: cotx.
Answers
Answered by
MathMate
Even though you are given two possible answers, you are expected to explain how your choice was made.
Whenever you post an expression involving a division, you will need to put parentheses around the numerator and around the denominator to ensure an unambiguous definition of the problem.
As is, the expression posted is interpreted as:
cot(-x) - (1/1) - tan(x)
because of the rule of priority of operations, namely multiplication and division before addition and subtraction.
I have to <i>assume</i> that the intended expression is
(cot(-x)-1) / (1-tan(-x))
First we apply the normal trigonometric identities of
cot(-x)=-cot(x) and
tan(-x)=-tan(x)
because both tan(x) and cot(x) are odd functions.
The expression now becomes:
(-cot(x)-1) / (1+tan(x))
As with most identities involving tan(x), you are likely to get somewhere by substituting tan(x)=sin(x)/cos(x) and similarly for cot(x).
-( (cos(x)+ sin(x))/sin(x) ) / ( cos(x)+sin(x))/cos(x) )
= -cos(x)/sin(x)
= -cot(x)
Whenever you post an expression involving a division, you will need to put parentheses around the numerator and around the denominator to ensure an unambiguous definition of the problem.
As is, the expression posted is interpreted as:
cot(-x) - (1/1) - tan(x)
because of the rule of priority of operations, namely multiplication and division before addition and subtraction.
I have to <i>assume</i> that the intended expression is
(cot(-x)-1) / (1-tan(-x))
First we apply the normal trigonometric identities of
cot(-x)=-cot(x) and
tan(-x)=-tan(x)
because both tan(x) and cot(x) are odd functions.
The expression now becomes:
(-cot(x)-1) / (1+tan(x))
As with most identities involving tan(x), you are likely to get somewhere by substituting tan(x)=sin(x)/cos(x) and similarly for cot(x).
-( (cos(x)+ sin(x))/sin(x) ) / ( cos(x)+sin(x))/cos(x) )
= -cos(x)/sin(x)
= -cot(x)
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