(tan/(1+sec)) + ((1+sec)/tan) = 2csc
(Show all work please.)
I see tangents and secants on the left side, so I suspect to use the identity
tan^2x + 1 = sec^2x
LS
= tanx(1+secx) + (1+secx)/tanx
= (tan^2x + 1 + 2secx + sec^2x)/(tanx(1+secx))
= (sec^2x + 2secx + sec^2x)/(tanx(1+secx))
= (2sec^2x + 2secx)/(tanx(1+secx))
= 2secx(secx + 1)/(tanx(1+secx))
= 2secx/tanx
= (2/cosx)÷(sinx/cosx)
= (2/cosx)(cosx/sinx)
= 2/sinx
= 2cscx
= RS
Q.E.D.
y=2sec(1/3x)
To solve the equation,
(tan/(1+sec)) + ((1+sec)/tan) = 2csc
First, let's simplify the expression on the left-hand side of the equation.
We can start by finding the reciprocal of the sine function:
csc(x) = 1/sin(x)
Therefore, we rewrite 2csc as:
2csc = 2(1/sin)
Next, let's simplify the expression (tan/(1+sec)).
We know that tan(x) = sin(x)/cos(x)
So, tan/(1+sec) becomes:
(sin/cos)/(1 + 1/cos)
Now, let's find a common denominator for the expression on the left side.
The common denominator for (1 + 1/cos) is cos.
So, we have:
(sin/cos)/(1 + 1/cos) = (sin/cos)/(cos/cos + 1/cos) = (sin/cos)/((1 + 1)/cos) = (sin/cos)/(2/cos) = sin/(2cos)
Similarly, let's simplify the expression ((1+sec)/tan).
We know that sec(x) = 1/cos(x).
Therefore, (1+sec)/tan becomes:
(1 + 1/cos)/(sin/cos) = (1 + (1/cos))/(sin/cos) = (cos/cos + 1/cos)/(sin/cos) = (cos + 1)/sin
Now, we can rewrite the equation:
(sin/(2cos)) + ((cos + 1)/sin) = 2(1/sin)
To eliminate the fractions, we can multiply through by their common denominator, sin(2cos):
sin/(2cos) * sin(2cos) + (cos + 1)/sin * sin(2cos) = 2(1/sin) * sin(2cos)
simplifying,
sin(sin(2cos)) + (cos + 1)(2cos) = 2(2cos)
Expanding the expressions,
sin(2cos) + 2cos^2 + 2cos + sin(2cos) = 4cos
Simplifying,
2sin(2cos) + 2cos^2 + 2cos - 4cos = 0
Combining like terms,
2sin(2cos) + 2cos^2 - 2cos = 0
Now, let's solve the equation.