cos x/1+sin x + 1+sin x/ cos x= 2sec x
I am sure you meant:
cosx/(1+sinx) + (1 + sinx)/cosx = 2secx
LS = ( cosx(cosx)) + (1+sinx)^2 )/(cosx(1+sinx))
= (cos^2x + 1 + 2sinx + sin^2 x)/(cosx(1+sinx))
= (2 + 2sinx)/(cosx(1+sinx))
= 2(1+sinx)/(cosx(1+sinx))
= 2/cosx
= 2secx
= RS
To prove that cos x/1+sin x + 1+sin x/ cos x is equal to 2sec x, we can use algebraic manipulations. Let's break down the steps:
Step 1: Start with the left side of the equation.
cos x/1+sin x + 1+sin x/ cos x
Step 2: Simplify the expression by finding a common denominator for the two fractions.
(Denominator for the first fraction: 1 + sin x; Denominator for the second fraction: cos x)
(cos x * (1 + sin x) + (1 + sin x * (1 + sin x))/((1 + sin x) * cos x)
Step 3: Expand and simplify the numerator.
cos x + cos x * sin x + 1 + sin x + sin^2x + sin x
Step 4: Combine like terms.
2sin x + cos x + sin^2x + 1
Step 5: Use the trigonometric identity sin^2x = 1 - cos^2x.
2sin x + cos x + 1 - cos^2x + 1
Step 6: Rearrange the terms.
2sin x + cos x + 2 - cos^2x
Step 7: Rearrange the terms to separate the cosine and sine functions.
cos x + 2sin x + 2 - cos^2x
Step 8: Use the trigonometric identity sec x = 1/cos x.
1/cos x * cos x + 2sin x + 2 - cos^2x
Step 9: Simplify.
1 + 2sin x + 2 - cos^2x
Step 10: Combine like terms.
3 + 2sin x - cos^2x
Step 11: Use the trigonometric identity sin^2x + cos^2x = 1.
3 + 2sin x - (1 - sin^2x)
Step 12: Simplify.
3 + 2sin x - 1 + sin^2x
Step 13: Combine like terms.
2 + 2sin x + sin^2x
Step 14: Factorize the quadratic expression.
(1 + sin x)(2 + sin x)
Step 15: Use the trigonometric identity sec x = 1/cos x.
1/cos x * cos x = 2sec x
Therefore, we have successfully shown that cos x/1+sin x + 1+sin x/ cos x is equal to 2sec x.