tanx +cotx=secx cscx
Yes, this seems to be a true statement.
start with the LS
LS = tanx + cotx
= sinx/cosx + cosx/sinx
your LCD is sinxcosx
= (sin^2 x + cos^2 x)/(sinxcosx)
= ....
you should be able to finish it
To prove the given equation, we need to simplify the left-hand side (LHS) of the equation and show that it is equal to the right-hand side (RHS).
Let's start by simplifying the LHS using trigonometric identities:
tan(x) + cot(x)
We know that:
tan(x) = sin(x)/cos(x)
cot(x) = cos(x)/sin(x)
Substituting these values, we get:
sin(x)/cos(x) + cos(x)/sin(x)
Now, to add these fractions, we need to find a common denominator. The common denominator in this case is cos(x) * sin(x). Therefore, we multiply the first fraction by sin(x)/sin(x) and the second fraction by cos(x)/cos(x):
(sin(x)*sin(x))/(cos(x)*sin(x)) + (cos(x)*cos(x))/(sin(x)*cos(x))
Simplifying further, we get:
sin^2(x)/(cos(x)*sin(x)) + cos^2(x)/(sin(x)*cos(x))
Now, let's simplify each part separately:
sin^2(x) can be written as 1 - cos^2(x) using the Pythagorean identity for sin^2(x) + cos^2(x) = 1.
Thus, sin^2(x) = 1 - cos^2(x).
Substituting this into our equation, we have:
(1 - cos^2(x))/(cos(x)*sin(x)) + cos^2(x)/(sin(x)*cos(x))
Now, combining the fractions, we get:
(1 - cos^2(x) + cos^2(x))/(cos(x)*sin(x))
Simplifying further:
1/(cos(x)*sin(x))
We know that:
1/sin(x) = csc(x)
1/cos(x) = sec(x)
Substituting these values, we finally have:
sec(x) * csc(x) = sec(x) * csc(x)
Therefore, we have proven that tan(x) + cot(x) = sec(x) * csc(x).