Prove that
tanx+cotx=2sec(2x)
Thank you! :)
LS = sinx/cosx + cosx/sinx
= (sin^2 x + cos^2 x)/sinxcosx
= 1/(sinxcosx)
= 2/(2sinxcosx)
= 2/sin (2x)
= 2csc (2x)
Your RS = 2sec(2x) which is incorrect
Did you have a typo?
Yes I had a typo. Thank you!!
(I don't know how to respond to answers besides this....)
To prove that tanx + cotx = 2sec(2x), we need to simplify both sides of the equation and show that they are equal.
Starting with the left-hand side (LHS), we can rewrite tanx + cotx using their definitions in terms of sinx and cosx:
LHS: tanx + cotx
= sinx/cosx + cosx/sinx
= (sin^2x + cos^2x)/(sinx*cosx) (common denominator)
= 1/(sinx*cosx) (trigonometric identity: sin^2x + cos^2x = 1)
= secx*(1/cosx) (multiplying by secx/secx)
= secx/cosx (reciprocal identity: 1/cosx = secx)
= 2secx*(1/2cosx) (multiplying by 2/2)
= 2secx*(1/(2cos^2x - 1 + 1)) (using cos^2x - 1 = -sin^2x + 1)
= 2secx*(1/(1 - sin^2x + 1)) (using 2cos^2x = 1 + cos2x)
= 2secx*(1/(2 - sin^2x))
Now, let's consider the right-hand side (RHS):
RHS: 2sec(2x)
To simplify further, recall that sec(2x) can be expressed as 1/cos(2x):
RHS: 2sec(2x)
= 2/(1/cos(2x))
= 2*cos(2x)
To compare the LHS and RHS, we need to express cos(2x) in terms of sinx:
cos(2x) = 1 - 2sin^2x (double-angle identity: cos(2x) = 1 - 2sin^2x)
Substituting this expression into the RHS:
RHS: 2*cos(2x)
= 2*(1 - 2sin^2x)
= 2 - 4sin^2x
Now, we have:
LHS: 2secx*(1/(2 - sin^2x))
RHS: 2 - 4sin^2x
Notice that 2secx*(1/(2 - sin^2x)) is equal to 2 - 4sin^2x because the expression (2 - sin^2x) is equal to (1 - 2sin^2x).
Therefore, we have proven that tanx + cotx = 2sec(2x).
To prove that tan(x) + cot(x) = 2sec(2x), we need to start with one side of the equation and manipulate it until we reach the other side.
We'll start with the right side of the equation 2sec(2x). Recall that the secant function is the reciprocal of the cosine function:
sec(x) = 1/cos(x)
So, we can rewrite 2sec(2x) as:
2/cos(2x)
Now, let's manipulate the left side of the equation, tan(x) + cot(x).
Recall that the tangent function is the ratio of the sine function to the cosine function:
tan(x) = sin(x)/cos(x)
And the cotangent function is the reciprocal of the tangent function:
cot(x) = 1/tan(x) = cos(x)/sin(x)
Now, let's substitute these values into the left side of the equation:
tan(x) + cot(x) = sin(x)/cos(x) + cos(x)/sin(x)
To add these two fractions, we need a common denominator. The common denominator is the product of the two denominators, so we multiply the first fraction by (sin(x)/sin(x)) and the second fraction by (cos(x)/cos(x)):
tan(x) + cot(x) = (sin(x)/sin(x))*(sin(x)/cos(x)) + (cos(x)/cos(x))*(cos(x)/sin(x))
Simplifying the fractions:
tan(x) + cot(x) = (sin²(x) + cos²(x))/(sin(x)*cos(x))
Since sin²(x) + cos²(x) = 1 for any angle x, we can substitute that in:
tan(x) + cot(x) = 1/(sin(x)*cos(x))
Now, let's manipulate this expression further to make it match the expression we obtained for the right side of the equation:
tan(x) + cot(x) = 1/(sin(x)*cos(x))
Multiplying the numerator and denominator by 2, we have:
tan(x) + cot(x) = 2 / (2*sin(x)*cos(x))
Using the identity sin(2x) = 2*sin(x)*cos(x), we can rewrite the expression as:
tan(x) + cot(x) = 2 / sin(2x)
Finally, recall that the reciprocal of the sine function is the cosecant function:
csc(x) = 1/sin(x)
Therefore, the expression becomes:
tan(x) + cot(x) = 2 * csc(2x)
We have now reached the left side of the equation:
tan(x) + cot(x) = 2 * csc(2x)
Which matches the right side of the equation:
2sec(2x)
Hence, we have proven that tan(x) + cot(x) = 2sec(2x).