prove that cosx(tanx+2)(2tanx+1)=2secx+5sinx
Good, but i can do better than this
Looks like "homework or assignment dumping" to me
Exactly what problems do you have with this list of trig questions?
I will do this one.... show me some attempts at the others
LS = cosx(2tan^2 x + 5tanx + 2)
= 2cosx sin^2 x/cos^2 x + 5cosxsinx/cosx + 2cosx
= 2sin^2 x/cosx + 5sinx + 2cosx
= (2sin^2 x + 5sinxcosx + 2cos^2 x)/cosx
= (2 + 5sinxcosx)/cosx
= 2/cosx + 5sinx
= 2secx + 5sinx
= RS
To prove the equation cos(x)(tan(x) + 2)(2tan(x) + 1) = 2sec(x) + 5sin(x), we need to simplify both sides of the equation and show that they are equal. Let's start by simplifying the left-hand side (LHS) of the equation.
LHS: cos(x)(tan(x) + 2)(2tan(x) + 1)
Step 1: Distribute the cos(x) to the terms inside the parentheses:
LHS = cos(x) * tan(x)(2tan(x) + 1) + cos(x)(2tan(x) + 1)(2)
Step 2: Simplify each term separately:
Term 1: cos(x) * tan(x)(2tan(x) + 1)
To simplify this term, we can use the identity tan(x) = sin(x)/cos(x).
Substituting, we get:
cos(x) * (sin(x)/cos(x))(2tan(x) + 1)
The cosine terms cancel out, and we are left with:
sin(x)(2tan(x) + 1)
Term 2: cos(x)(2tan(x) + 1)(2)
Expanding this term, we have:
cos(x) * 2 * (2tan(x) + 1)
Simplifying further, we get:
4cos(x)(tan(x) + 1)
Now, let's combine both terms together:
LHS = sin(x)(2tan(x) + 1) + 4cos(x)(tan(x) + 1)
Next, let's simplify the right-hand side (RHS) of the equation:
RHS = 2sec(x) + 5sin(x)
Step 3: Substitute sec(x) with its reciprocal:
Recall that sec(x) is the reciprocal of cos(x), so we can replace sec(x) with 1/cos(x).
RHS = 2(1/cos(x)) + 5sin(x)
Multiplying both terms by cos(x), we get:
RHS = 2 + (5sin(x)cos(x))/cos(x)
The sine and cosine terms in the numerator (5sin(x)cos(x)) cancel out, and we are left with:
RHS = 2 + 5sin(x)
Now, let's compare the simplified LHS and RHS to see if they are equal.
LHS = sin(x)(2tan(x) + 1) + 4cos(x)(tan(x) + 1)
= sin(x)(2sin(x)/cos(x) + 1) + 4cos(x)(sin(x)/cos(x) + 1)
= sin(x)(2sin(x) + cos(x))/cos(x) + 4cos(x)(sin(x) + cos(x))/cos(x)
= (2sin(x)sin(x) + sin(x)cos(x))/cos(x) + (4sin(x)cos(x) + 4cos(x)cos(x))/cos(x)
= (2sin(x)^2 + sin(x)cos(x))/cos(x) + (4sin(x)cos(x) + 4cos(x)^2)/cos(x)
= (sin(x)(2sin(x) + cos(x)) + 4(sinx + cos(x))^2)/cos(x)
RHS = 2 + 5sin(x)
We can see that both sides are not equal to each other. Therefore, the initial equation:
cos(x)(tan(x) + 2)(2tan(x) + 1) = 2sec(x) + 5sin(x)
is not true.