Verify
cos x/1+sinx + 1+ sinx/cosx = 2sec
in this case you must use brackets or else the statement is false
LS = cosx/(1+sinx) + (1+sinx)/cosx
= cosx/(1+sinx) + (1+sinx)/cosx
= ( cosx(cosx) + (1+sinx)(1+sinx) )/(cosx(1+sinx))
= (cos^2 x + 1 + 2sinx + sin^2 x)/(cosx(1+sinx))
= (1 + 1 + 2sinx)/(cosx(1+sinx))
= 2(1+sinx)/(cosx(1+sinx))
= 2/cosx
= 2secx
= RS
To verify the given equation, we need to simplify both sides of the equation and show that they are equal.
Given:
cos x/(1 + sin x) + (1 + sin x)/cos x = 2sec x
To start, let's simplify the left-hand side of the equation.
cos x/(1 + sin x) + (1 + sin x)/cos x
To add these fractions, we need to find a common denominator. The common denominator is (1 + sin x)(cos x).
(cos x * cos x)/(cos x * (1 + sin x)) + (1 + sin x)(1 + sin x)/(cos x * (1 + sin x))
Now, let's simplify the numerators:
cos^2 x + (1 + sin x)(1 + sin x)
Expanding (1 + sin x)(1 + sin x):
cos^2 x + (1 + 2sin x + sin^2 x)
Combining the terms:
cos^2 x + 1 + 2sin x + sin^2 x
Next, let's simplify the denominator:
cos x * (1 + sin x)
Now, let's simplify further:
cos^2 x + 1 + 2sin x + sin^2 x / cos x * (1 + sin x)
Using the identity sin^2 x + cos^2 x = 1, we can replace cos^2 x with 1 - sin^2 x:
(1 - sin^2 x) + 1 + 2sin x + sin^2 x / cos x * (1 + sin x)
Simplifying further:
1 - sin^2 x + 1 + 2sin x + sin^2 x / cos x * (1 + sin x)
The sin^2 x and -sin^2 x terms cancel out, and we have:
2 + 2sin x / cos x * (1 + sin x)
Now, let's simplify the right-hand side of the equation:
2sec x
Recall that sec x is the reciprocal of cos x:
2(1/cos x)
Simplifying:
2/cos x
Now, our equation becomes:
2 + 2sin x / cos x * (1 + sin x) = 2/cos x
To make the equation easier to work with, let's multiply both sides of the equation by cos x:
2cos x + 2sin x(1 + sin x) = 2
Distributing through the right-hand side:
2cos x + 2sin x + 2sin^2 x = 2
On the left-hand side, let's combine like terms:
2sin^2 x + 2sin x + 2cos x = 2
Since sin^2 x + cos^2 x = 1, we can replace sin^2 x with 1 - cos^2 x:
2(1 - cos^2 x) + 2sin x + 2cos x = 2
Simplifying further:
2 - 2cos^2 x + 2sin x + 2cos x = 2
Combining like terms:
2sin x + 2cos^2 x - 2cos x = 0
Now, let's factor out a 2 from the left-hand side:
2(sin x + cos^2 x - cos x) = 0
Dividing both sides of the equation by 2:
sin x + cos^2 x - cos x = 0
Using the identity cos^2 x = 1 - sin^2 x, we can replace cos^2 x with 1 - sin^2 x:
sin x + (1 - sin^2 x) - cos x = 0
Distributing through the right-hand side:
sin x + 1 - sin^2 x - cos x = 0
Rearranging the terms:
-sin^2 x + sin x - cos x + 1 = 0
Since -sin^2 x = -(1 - cos^2 x), we can substitute -(1 - cos^2 x) for -sin^2 x:
-(1 - cos^2 x) + sin x - cos x + 1 = 0
Expanding -(1 - cos^2 x):
-1 + cos^2 x + sin x - cos x + 1 = 0
The -1 and +1 cancel out, and we have:
cos^2 x + sin x - cos x = 0
Using the identity cos x - sin x = sin(π/2 - x), we can substitute sin(π/2 - x) for cos x - sin x:
cos^2 x + sin x - sin(π/2 - x) = 0
Rearranging the terms:
cos^2 x + sin(π/2 - x) + sin x = 0
Now, using the identity cos^2 x = 1 - sin^2 x, we can replace cos^2 x with 1 - sin^2 x:
1 - sin^2 x + sin(π/2 - x) + sin x = 0
Expanding sin(π/2 - x) as cos x:
1 - sin^2 x + cos x + sin x = 0
The 1 and 0 cancel out:
-sin^2 x + cos x + sin x = 0
Using the identity sin x - sin^2 x = sin x(1 - sin x), we can factor out sin x:
sin x(1 - sin x) + cos x = 0
This equation is different than the original equation, so it seems that there was a mistake made in the simplification process.
Hence, the verification of the given equation is not valid.