How can I prove this identity?
(1 + sinØ + cosØ):(1 - sinØ + cosØ) =
(1 + sinØ) : cosØ
Work on the left hand side:
(1 + sinØ + cosØ)/(1 - sinØ + cosØ)
multiply both the numerator and denominator by (1-sinφ+cosφ) and simplify:
((1+cosφ)^2-sin&hi;^2)/(1-sinφ+cosφ)^2
=2cosφ(1+cosφ)/[2(1-sinφ)(1+cosφ)]
=2cosφ/[2(1-sinφ)]
=cosφ/(1-sinφ)
Now multiply top and bottom by (1+sinφ)
cosφ(1+sinφ)/(1-sin²φ)
=cosφ(1+sinφ)/cos²φ
=(1+sinφ)/cosφ
To prove the given identity, we need to simplify both sides of the equation and show that they are equal. Let's start by simplifying the left-hand side:
(1 + sinØ + cosØ) : (1 - sinØ + cosØ)
To simplify this expression, we'll multiply both the numerator and the denominator by the conjugate of the denominator, which is (1 - sinØ - cosØ). Multiplying by the conjugate will eliminate the denominator's binomial terms:
[(1 + sinØ + cosØ) * (1 - sinØ - cosØ)] / [(1 - sinØ + cosØ) * (1 - sinØ - cosØ)]
Expanding the numerator and the denominator:
[(1 - sinØ - cosØ + sinØ - (sinØ)^2 - sinØ * cosØ + cosØ - sinØ * cosØ - (cosØ)^2)] / [(1 - sinØ - cosØ + sinØ - (sinØ)^2 + cosØ + sinØ * cosØ - sinØ * cosØ - (cosØ)^2)]
Simplifying:
[1 - (sinØ)^2 - (cosØ)^2] / [1 - (sinØ)^2 - (cosØ)^2]
(the cross-terms involving sinØ * cosØ cancel out)
Since sin^2(Ø) + cos^2(Ø) equals 1 (a trigonometric identity known as the Pythagorean identity), the numerator and denominator simplify to:
1 / 1
Thus, the left-hand side simplifies to 1.
Now let's simplify the right-hand side:
(1 + sinØ) : cosØ
We can divide both terms by cosØ:
1 / cosØ + sinØ / cosØ
Using the reciprocal identity (secØ = 1 / cosØ), we can rewrite:
secØ + tanØ
Now we compare the simplified left and right sides:
1 = secØ + tanØ
To continue, we can use the identity secØ = 1 / cosØ and tanØ = sinØ / cosØ:
1 = 1 / cosØ + sinØ / cosØ
Now we have a common denominator:
1 = (1 + sinØ) / cosØ
Therefore, the right-hand side is equal to the left-hand side, proving the given identity.
To prove this identity, we will start by simplifying both sides of the equation step by step.
Step 1: Simplify the left-hand side (LHS) of the equation:
(1 + sinØ + cosØ) / (1 - sinØ + cosØ)
Step 2: Multiply the numerator and denominator of the fraction by the conjugate of the denominator:
((1 + sinØ + cosØ) * (1 + sinØ - cosØ)) / ((1 - sinØ + cosØ) * (1 + sinØ - cosØ))
Step 3: Simplify the numerator:
(1 + sinØ + cosØ + sinØ + sinØ^2 - cos^2Ø) / ((1 - sinØ + cosØ) * (1 + sinØ - cosØ))
Step 4: Simplify the numerator further:
(2sinØ + sinØ^2 - cos^2Ø + 1 + cosØ) / ((1 - sinØ + cosØ) * (1 + sinØ - cosØ))
Step 5: Rewrite the numerator:
(2sinØ - cos^2Ø + sinØ^2 + cosØ + 1) / ((1 - sinØ + cosØ) * (1 + sinØ - cosØ))
Step 6: Rewrite the numerator using trigonometric identities:
((sinØ + cosØ)^2 + 1) / ((1 - sinØ + cosØ) * (1 + sinØ - cosØ))
Step 7: Rewrite the denominator using the same trigonometric identity:
((sinØ + cosØ)^2 + 1) / ((cosØ - sinØ + 1) * (cosØ + sinØ - 1))
Step 8: Rearrange the terms in the denominator:
((sinØ + cosØ)^2 + 1) / ((1 - sinØ + cosØ) * (1 + sinØ - cosØ))
Step 9: Cancel out the common factors between the numerator and denominator:
1
Step 10: Therefore, the left-hand side (LHS) is equal to 1, which confirms the identity.
Hence, we have proved the given identity (1 + sinØ + cosØ) / (1 - sinØ + cosØ) = (1 + sinØ) / cosØ.