how do you solve this problem?
1+sinØ/cosØ +cosØ/1+sinØ = 2secØ
and
cosß- cosß/1-tanß = sinßcosß/sinß-cosß
for your second question you probably meant (I will switch ß to x, easier to type)
cosß- cosß/(1-tanß) = sinßcosß/(sinß-cosß)
Left Side
= cosx - cosx/(1-sinx/cosx)
= cosx - cosx/[(cosx-sinx)/cosx]
=cosx - cos^2 x/(cosx-sinx)
=(cos^2 x - sinxcosx - cos^2 x)/cosx - sinx)
= -sinxcosx/(cosx-sinx)
= sinxcosx/(sinx-cosx)
= Right Side
To solve the first problem:
Step 1: Simplify the left-hand side of the equation.
Using the identities sin²Ø + cos²Ø = 1 and secØ = 1/cosØ:
1 + sinØ/cosØ + cosØ/1 + sinØ = (1/cosØ) + (cosØ / (1 + sinØ))
= (1 + cosØ) / cosØ + cosØ / (1 + sinØ)
= (1 + cosØ + cosØ) / cosØ + cosØ / (1 + sinØ)
= (1 + 2cosØ) / cosØ + cosØ / (1 + sinØ)
Step 2: Combine the fractions on the left-hand side into a single fraction.
Using a common denominator of cosØ (1 + sinØ):
[(1 + 2cosØ)(1 + sinØ) + cosز] / [cosØ(1 + sinØ)]
Step 3: Expand and simplify the numerator.
[(1 + 2cosØ + sinØ + 2sinØcosØ) + cosز] / [cosØ(1 + sinØ)]
= [1 + sinØ + cosØ + 2cosز + 2sinØcosØ] / [cosØ(1 + sinØ)]
Step 4: Simplify further if possible.
Combine like terms in the numerator:
= [3cosØ + sinØ + 2cosز + 2sinØcosØ] / [cosØ(1 + sinØ)]
Step 5: Apply the identity cos²Ø = 1 - sin²Ø to simplify the numerator.
Use the distributive property to expand 2sinØcosØ:
= [3cosØ + sinØ + 2cosز + sinØ(1 - cos²Ø)] / [cosØ(1 + sinØ)]
= [3cosØ + sinØ + 2cosز + sinØ - sinØcos²Ø] / [cosØ(1 + sinØ)]
= [4cosØ + 2sinØ - sinØcos²Ø + 2cosز] / [cosØ(1 + sinØ)]
Step 6: Simplify further if possible.
Rearrange and factor out common terms:
= [2(cosØ + sinØ) + (2cosز - sinØcos²Ø)] / [cosØ(1 + sinØ)]
Step 7: Apply the identity 1 + sinØ = secØ to simplify the denominator.
Using the fact that secØ = 1/cosØ:
= [2(cosØ + sinØ) + (2cosز - sinØcos²Ø)] / [cosØ * secØ]
= [2(cosØ + sinØ) + (2cosز - sinØcos²Ø)] / [cosØ / cosØ]
= 2(cosØ + sinØ) + (2cosز - sinØcos²Ø)
Step 8: Simplify further if possible.
Expand the square term using the identity cos²Ø = 1 - sin²Ø:
= 2(cosØ + sinØ) + [2cosز - sinØ(1 - sin²Ø)]
= 2(cosØ + sinØ) + [2cosز - sinØ + sinز]
= 2(cosØ + sinØ) + (2cosز - sinØ + sinز)
= 2(cosØ + sinØ) + (2cosز + sinز - sinØ)
Step 9: Simplify further if possible.
Combine like terms in the equation:
2(cosØ + sinØ) + (2cosز + sinز - sinØ) = 2secØ
2cosØ + 2sinØ + 2cosز + sinز - sinØ = 2secØ
2cosØ + 2sinØ + 2cosز + sinز - sinØ - 2secØ = 0
This equation can be further simplified by factoring, but it is not possible to solve it. The equation contains both trigonometric functions and a secant term, which makes it difficult to find an exact solution.
To solve the second problem:
Let's simplify the expression step by step.
Step 1: Simplify the left-hand side of the equation.
cosß - cosß / (1 - tanß) = sinßcosß / (sinß - cosß)
Step 2: Combine the fractions on each side.
(cosß(1 - cosß)) / (1 - tanß) = (sinßcosß) / (sinß - cosß)
Step 3: Apply the identity cos²ß = 1 - sin²ß to simplify the numerator on the left side.
(cosß(1 - cosß)) / (1 - tanß) = (sinßcosß) / (sinß - cosß)
(cosß - cos²ß) / (1 - tanß) = (sinßcosß) / (sinß - cosß)
Step 4: Simplify further if possible.
Using the identites:
sinß = cos(π/2 - ß)
cos²ß = 1 - sin²ß
We can rewrite the equation as:
(cosß - (1 - sin²ß)) / (1 - tanß) = sinßcosß / (sinß - cosß)
([cosß - 1 + sin²ß]) / (1 - tanß) = [cos(π/2 - ß)]cosß / [cos(π/2 - ß) - cosß]
([sin²ß - cosß + 1]) / (1 - tanß) = [cos(π/2 - ß)]cosß / [sin(π/2 - ß) - sinß]
Step 5: Simplify further if possible.
Using the identity cos(π/2 - A) = sinA and sin(π/2 - A) = cosA, the equation becomes:
([sin²ß - cosß + 1]) / (1 - tanß) = (cosß)cosß / (cos(ß) - sinß)
Step 6: Simplify further if possible.
Multiply both sides of the equation by (1 - tanß)(cosß - sinß) to eliminate the denominators:
([sin²ß - cosß + 1])(cosß - sinß) = (cosß)cosß(1 - tanß)
Step 7: Expand and simplify the equation.
(sin²ßcosß - cos²ßsinß + cosß - sin²ß + cosßsinß - sin²ßsinß) = (cos²ß - cos²ßtanß)
Step 8: Combine like terms and simplify.
(sin²ßcosß - 2cos²ßsinß + 2cosß - 2sin²ß) = (cos²ß - cos²ßtanß)
At this point, the equation cannot be further simplified.
To solve the first problem, let's simplify the expression step by step:
1+sinØ/cosØ +cosØ/1+sinØ = 2secØ
Step 1: Simplify the left-hand side (LHS) by finding a common denominator for the fractions:
[(1+sinØ)(1+sinØ)]/(cosØ)(1+sinØ) + [(cosØ)(cosØ)]/(cosØ)(1+sinØ) = 2secØ
Step 2: Combine the fractions on the LHS:
[(1+sinØ)(1+sinØ) + (cosØ)(cosØ)] / [(cosØ)(1+sinØ)] = 2secØ
Step 3: Expand and simplify the numerator on the LHS:
[1+2sinØ+sin²Ø + cos²Ø] / [(cosØ)(1+sinØ)] = 2secØ
Step 4: Simplify the numerator by using the identity sin²Ø + cos²Ø = 1:
[2+2sinØ] / [(cosØ)(1+sinØ)] = 2secØ
Step 5: Divide both sides by 2:
[1+sinØ] / [(cosØ)(1+sinØ)] = secØ
Step 6: Cancel out the common factor (1+sinØ) on the top and bottom:
1 / cosØ = secØ
Since 1 / cosØ is equal to secØ, the equation simplifies to:
secØ = secØ
Therefore, the first problem has infinitely many solutions. Any angle Ø that satisfies the equation secØ = secØ will be a solution.
For the second problem:
cosß- cosß/1-tanß = sinßcosß/sinß - cosß
Step 1: Simplify the left-hand side (LHS) by finding a common denominator for the fractions:
[(cosß)(1-tanß) - cosß]/(1-tanß) = sinßcosß / [sinß - cosß]
Step 2: Expand and simplify the numerator on the LHS:
[cosß - cosßtanß - cosß]/(1-tanß) = sinßcosß / [sinß - cosß]
Step 3: Combine like terms on the LHS:
-cosßtanß/(1-tanß) = sinßcosß / [sinß - cosß]
Step 4: Multiply both sides by (1-tanß) to eliminate the fraction on the LHS:
-cosßtanß = [(sinßcosß)(1-tanß)] / [sinß - cosß]
Step 5: Cancel out the common factor (sinßcosß) on the top and bottom:
-cosßtanß = 1 - tanß
Step 6: Add cosßtanß to both sides:
0 = 1
Since 0 does not equal 1, the second problem has no solution.
Since you did not use parentheses for your numerators and denominators, I spent most of my time figuring out what you probably mean
(1 + sin x)/( cos x) + cos x/(1+sin x) = 2/cos x
common denominator on left of
(1+sin x) cos x
[(1 +sin x)^2 +cos^2 x ]/common denom
[ 1 + 2 sin x + sin^2 x +1 - sin^2 x ] / common denom
[ 2 + 2 sin x ] / [(1+sin x) cos x ]
I think you can take it from there