Prove This trig identity cos/1-tan +Sin/1-cot = Sin +Cos
To prove the identity, we need to manipulate the left-hand side (LHS) until it matches the right-hand side (RHS).
Starting with the LHS:
cos/(1 - tan) + sin/(1 - cot)
To simplify, we'll express tan and cot in terms of sin and cos.
Recall that tan(theta) = sin(theta) / cos(theta) and cot(theta) = cos(theta) / sin(theta).
So, tan = sin / cos and cot = cos / sin.
Substituting these values into the LHS:
cos/(1 - (sin/cos)) + sin/(1 - (cos/sin))
Now, let's simplify the expressions inside the denominators:
1 - (sin/cos) = (cos - sin) / cos
1 - (cos/sin) = (sin - cos) / sin
Substituting these values back into the LHS:
cos/((cos - sin)/cos) + sin/((sin - cos)/sin)
To simplify further, we can multiply both numerator and denominator of each fraction by their respective denominators:
cos * (cos / (cos - sin)) + sin * (sin / (sin - cos))
Now, we can cancel out the common factors:
(cos^2) / (cos - sin) + (sin^2) / (sin - cos)
Next, we'll try to combine the fractions by finding a common denominator. To do this, we'll multiply the first fraction by (sin - cos) / (sin - cos) and the second fraction by (cos - sin) / (cos - sin):
[(cos^2 * (sin - cos)) + (sin^2 * (cos - sin))] / [(cos - sin)(sin - cos)]
Expanding the numerators:
[(cos^2sin - cos^3) + (sin^2cos - sin^3)] / [(cos - sin)(sin - cos)]
Now, we can collect like terms:
[(cos^2sin + sin^2cos) - (cos^3 + sin^3)] / [(cos - sin)(sin - cos)]
Again, note that cos^2sin + sin^2cos = sin(cos^2 + sin^2) = sin.
Substituting this simplification:
[sin - (cos^3 + sin^3)] / [(cos - sin)(sin - cos)]
Next, we'll use the identities cos^3 + sin^3 = (cos + sin)(cos^2 - cos sin + sin^2) and cos^2 - cos sin + sin^2 = 1.
So, cos^3 + sin^3 = (cos + sin)(1) = cos + sin.
Substituting this simplification:
[sin - (cos + sin)] / [(cos - sin)(sin - cos)]
Rearranging the numerator:
[(-cos) / (cos - sin)] / [(cos - sin)(sin - cos)]
Canceling out the common factor:
-1 / (cos - sin)
Now, let's rewrite the denominator as -(sin - cos):
-1 / (-(sin - cos))
Which simplifies to:
1 / (sin - cos) = sin + cos
Therefore, the LHS is equal to the RHS, and we have proven the trigonometric identity:
cos/(1 - tan) + sin/(1 - cot) = sin + cos.
To prove the trigonometric identity:
cos/(1-tan) + sin/(1-cot) = sin + cos
First, let's simplify the left-hand side of the equation:
cos/(1-tan) + sin/(1-cot)
To simplify this expression, we need to express both cos and sin in terms of tan and cot, respectively.
Recall the following identities:
cos^2 θ = 1 - sin^2 θ
cot θ = 1/tan θ
We'll use these identities to rewrite the trig functions:
cos = √(1 - sin^2)
tan = sin/cos => cos = sin/tan
cot = 1/tan
Now, let's substitute these expressions into the left-hand side of the equation:
√(1 - sin^2)/[1 - (sin/tan)] + sin/[1 - (1/tan)]
To simplify the expression further, let's find a common denominator:
√(1 - sin^2)/[(tan - sin)/tan] + sin/[(tan - 1)/tan]
To combine these fractions, we need to multiply the numerator and denominator of the first fraction by tan:
[√(1 - sin^2) * tan]/(tan - sin) + sin/[(tan - 1)/tan]
Now we have two fractions with a common denominator, so let's combine them:
[√(1 - sin^2) * tan + sin] / (tan - sin)
We can simplify the numerator by using the identity cos^2 θ = 1 - sin^2 θ:
[√(cos^2 - sin^2) * tan + sin] / (tan - sin)
Simplifying the square root:
[cos * tan + sin] / (tan - sin)
Now, let's simplify the denominator. We can use the identity cot θ = 1/tan θ:
[tan * (1/cot) + sin] / (tan - sin)
Simplifying the expression:
[(tan/cot) + sin] / (tan - sin)
Using the identity cot θ = 1/tan θ:
[(1/tan) + sin] / (tan - sin)
Finally, let's simplify the left-hand side:
[1 + tan*sin] / (tan - sin)
Now, let's simplify the right-hand side of the equation:
sin + cos
So, the given trigonometric identity:
cos/(1-tan) + sin/(1-cot) = sin + cos
is equivalent to the simplified expression:
[1 + tan*sin] / (tan - sin) = sin + cos
Since the left-hand side is equal to the right-hand side, the trigonometric identity is proved.