Exercise
$\frac{5csc^2x+4cscx-1}{\cot^2}=\frac{5csc\:x-1}{csc\:x\:-1}$
Step-by-step Solution
Learn how to solve factor by difference of squares problems step by step online. Prove the trigonometric identity (5csc(x)^2+4csc(x)+-1)/(cot(x)^2)=(5csc(x)-1)/(csc(x)-1). Starting from the left-hand side (LHS) of the identity. We can try to factor the expression 5\csc\left(x\right)^2+4\csc\left(x\right)-1 by applying the following substitution. Substituting in the polynomial, the expression results in. Factor the trinomial 5u^2+4u-1 of the form ax^2+bx+c, first, make the product of 5 and -1.
Prove the trigonometric identity (5csc(x)^2+4csc(x)+-1)/(cot(x)^2)=(5csc(x)-1)/(csc(x)-1)
Final answer to the exercise
true