(1/tanx-secx))+(1/(tanx+secx))=-2tanx
i've already attempted this question but something went wrong...
1/(tan-sec) + 1/(tan+sec)
= [(tan+sec)+(tan-sec)] / (tan^2 - sec^2)
= 2tan/(-1)
= -2tan
To solve this equation, we can start by simplifying the left-hand side (LHS) of the equation.
The LHS of the equation is:
(1/tanx - secx) + (1/(tanx + secx))
To simplify this expression, we need to find a common denominator for the two fractions.
The common denominator for tan(x) and sec(x) is (tan(x) + sec(x)). Multiplying the first fraction by (tan(x) + sec(x))/(tan(x) + sec(x)) and the second fraction by (tan(x) - sec(x))/(tan(x) - sec(x)), we get:
[(1 * (tan(x) + sec(x)))/(tan(x) + sec(x))] + [(1 * (tan(x) - sec(x)))/(tan(x) - sec(x))]
Simplifying further, we have:
[(tan(x) + sec(x))/(tan(x) + sec(x))] + [(tan(x) - sec(x))/(tan(x) - sec(x))]
Now we can combine the fractions:
(tan(x) + sec(x) + tan(x) - sec(x))/(tan(x) + sec(x) - tan(x) - sec(x))
Simplifying even further, we get:
(2tan(x))/(0)
At this point, we notice that the denominator is zero, which means the expression is undefined.
Since there is no solution where the equation holds true, we can conclude that there are no values of x that satisfy the given equation.