Prove that tan (Lambda) cos^2 (Lambda)+sin^2 (Lamda)/sin(Lambda) = cos (Lambda) + sin (Lambda)
(tan*cos^2 + sin^2)/sin
(sin*cos + sin^2)/sin
sin(cos+sin)/sin
cos+sin
This is not that question which I ask
To prove the given trigonometric equation, we need to simplify both sides of the equation and show that they are equal.
Starting with the left side of the equation:
tan(λ) * cos^2(λ) + sin^2(λ) / sin(λ)
First, let's simplify the expression using trigonometric identities:
Recall that:
tan(λ) = sin(λ) / cos(λ)
Substituting this into the left side of the equation:
(sin(λ) / cos(λ)) * cos^2(λ) + sin^2(λ) / sin(λ)
Now simplify further:
(sin(λ) * cos^2(λ)) / cos(λ) + sin^2(λ) / sin(λ)
The first term, (sin(λ) * cos^2(λ)) / cos(λ), can be simplified by canceling out the common factors of sin(λ) and cos(λ):
sin(λ) * cos(λ)
The second term, sin^2(λ) / sin(λ), simplifies to sin(λ):
sin(λ)
Now, we can rewrite the left side of the equation as:
sin(λ) * cos(λ) + sin(λ)
Factoring out sin(λ) from both terms:
sin(λ) * (cos(λ) + 1)
Now, let's simplify the right side of the equation:
cos(λ) + sin(λ)
As we can see, the right side matches the simplified left side of the equation:
sin(λ) * (cos(λ) + 1)
Therefore, we have proven that:
tan(λ) * cos^2(λ) + sin^2(λ) / sin(λ) = cos(λ) + sin(λ)