sinØ (1+ tanØ) + cosØ(1+ cotØ) = (secØ + cosecØ)
Nice Question
Answer is very easy
some one has already solved...........
sinØ (1+ tanØ) + cosØ(1+ cotØ)
sinØ(sinØ+cosØ)/cosØ + cosØ(sinØ+cosØ)/sinØ
tanØ(sinØ+cosØ) + cotØ(sinØ+cosØ)
(tanØ+cotØ)(sinØ+cosØ)
(sin^2Ø+cos^2Ø)/(sinØcosØ) (sinØ+cosØ)
(sinØ+cosØ)/(sinØcosØ)
1/cosØ + 1/sinØ
secØ+cscØ
Good
To prove the identity sinØ (1+ tanØ) + cosØ(1+ cotØ) = secØ + cosecØ, we need to simplify both sides of the equation using trigonometric identities.
Let's start with the left-hand side (LHS):
sinØ (1+ tanØ) + cosØ(1+ cotØ)
First, let's simplify sinØ(1+ tanØ):
sinØ (1+ tanØ) = sinØ + sinØ * tanØ
Now, let's simplify cosØ(1+ cotØ):
cosØ(1+ cotØ) = cosØ + cosØ * cotØ
Now, we can rewrite the equation using these simplified expressions:
sinØ + sinØ * tanØ + cosØ + cosØ * cotØ
Now, let's simplify the right-hand side (RHS):
secØ + cosecØ
Using the reciprocal identities, we can write secØ as 1/cosØ, and cosecØ as 1/sinØ:
1/cosØ + 1/sinØ
To simplify this further, we need a common denominator:
(sinØ + cosØ) / (sinØ * cosØ)
Now, we can see that the right-hand side (RHS) expression can be simplified to:
(sinØ + cosØ) / (sinØ * cosØ)
Comparing the left-hand side (LHS) and the right-hand side (RHS), we can see that they are equivalent.
Therefore, the identity sinØ (1+ tanØ) + cosØ(1+ cotØ) = secØ + cosecØ is proven.
I assume you are proving this identity to be true.
What have you done so far?