Hi. I am reposting this. Can you verify. Thanks.
(cosθ) / (sinθ – 1) + (cosθ)/ (1 + sinθ) = -2tanθ
(1 / cos^2θ) – (2) (1/cos^2θ) (sinθ / cosθ) + (sin^2θ / cos^2θ) = (1 - sinθ) / (1 + sinθ)
ok
I usually start with the more complicated looking side and try to change it to the more simplified
LS = cosØ/(sinØ - 1) + cosØ(sinØ + 1) , note it didn't matter to change the order of that last denominator
common denominator of (sinØ-1) and (sinØ+1) is (sinØ-1)(sinØ+1)
ahhh, difference of squares!
result is sin^2 Ø -1 or -cos^2 Ø
= ( cosØ(sinØ - 1) + cosØ(sinØ + 1) )/((sinØ-1)(sinØ+1))
= (sinØcosØ - cosØ + sinØcosØ + cosØ)/-cos^2 Ø
= 2sinØcosØ/-cos^2 Ø
= -2sinØ/cosØ
= -2tanØ
= RS
the 2nd is a variation of one I did yesterday, let me try to find it
from last night at 10:06 pm
RS = (1-sinØ)/(1+ sinØ)
= (1-sinØ)/(1+ sinØ) * (1-sinØ)/(1- sinØ)
= (1 - 2sinØ + sin^2 Ø)/(1 - sin^2 Ø)
= (1 - 2sinØ + sin^2 Ø)/cos^2 Ø
= 1/cos^2 Ø - 2sinØ/cos^2 Ø + sin^2 Ø/cos^2Ø
= sec^2 Ø - 2(sinØ/cosØ)*(1/cosØ) + tan^2 Ø
= sec^2 Ø - 2tanØ secØ + tan^2 Ø
= 1/cos^2 Ø -2(sinØ/cosØ)(1/cosØ) + sin^2 Ø/cos^2 Ø
= your LS
Steve also gave you an alternate solution in that same post
To verify the equations, we can simplify both sides and check if they are equal.
Starting with the first equation:
(cosθ) / (sinθ – 1) + (cosθ) / (1 + sinθ) = -2tanθ
To simplify this expression, we can find a common denominator:
[(cosθ)(1 + sinθ) + (cosθ)(sinθ – 1)] / [(sinθ – 1)(1 + sinθ)]
Expanding and combining like terms in the numerator, we get:
[cosθ + cosθsinθ + cosθsinθ – cosθ] / [(sinθ – 1)(1 + sinθ)]
The two terms with cosθ cancel out, leaving:
2cosθsinθ / [(sinθ – 1)(1 + sinθ)]
Now, we can simplify the denominator as a difference of squares and express the expression in terms of tanθ:
2cosθsinθ / [(sinθ)^2 - 1]
Recall the trigonometric identity: sin^2θ + cos^2θ = 1
Rearranging, we get: cos^2θ = 1 - sin^2θ
Substituting this in our expression, we have:
2sinθcosθ / [cos^2θ - sin^2θ]
Now, using the trigonometric identity: tanθ = sinθ / cosθ
We can rewrite the expression as:
2tanθ
Therefore, the left side of the equation simplifies to -2tanθ, which matches the right side. Hence, the first equation is verified.
Moving on to the second equation:
(1 / cos^2θ) – (2) (1 / cos^2θ) (sinθ / cosθ) + (sin^2θ / cos^2θ) = (1 - sinθ) / (1 + sinθ)
To simplify this expression, we can combine the terms on the left side:
(1 - 2sinθ) / cos^2θ + (sin^2θ / cos^2θ) = (1 - sinθ) / (1 + sinθ)
Common denominator: cos^2θ
(1 - 2sinθ + sin^2θ) / cos^2θ = (1 - sinθ) / (1 + sinθ)
Now, we can simplify both the numerator and denominator using the identity: sin^2θ + cos^2θ = 1
(1 - sinθ)^2 / cos^2θ = (1 - sinθ) / (1 + sinθ)
Next, we can cancel out the common factor of (1 - sinθ):
(1 - sinθ) / cos^2θ = 1 / (1 + sinθ)
To simplify the denominator, we use the identity: cos^2θ = 1 - sin^2θ
(1 - sinθ) / (1 - sin^2θ) = 1 / (1 + sinθ)
Now, we can multiply both sides by (1 - sin^2θ):
(1 - sinθ)(1 + sinθ) = (1 - sin^2θ)
Expanding both sides:
1 - sin^2θ + sinθ - sin^3θ = 1 - sin^2θ
Simplifying:
-sin^3θ + sinθ = 0
Factoring out sinθ:
sinθ (1 - sin^2θ) = 0
Using the property of zero products, we get:
sinθ = 0 or 1 - sin^2θ = 0
From the first equation, we have sinθ = 0, which means θ = 0, π, 2π, etc.
From the second equation, we have 1 - sin^2θ = 0
Rearranging, we get sin^2θ = 1
Taking the square root of both sides, we have:
sinθ = ±1
So, θ can be π/2, 3π/2, etc.
Therefore, the solutions to the second equation are:
θ = 0, π, 2π, π/2, 3π/2, etc.
By verifying both equations, we can conclude if they are true or not.