Question

How do you verify the identity $\tan \dfrac{u}{2}=\cos ecu-\cot u?$

Answer 1

We will use some trigonometric identities to prove the given trigonometric identity. We will use the trigonometric identity given by $\tan \dfrac{x}{2}=\dfrac{1-\cos x}{\sin x}.$ We will apply this on the left-hand side of the equation. With some rearrangements, we can equate the left-hand side of the equation to the right-hand side of the equation.

Complete step by step solution:
Let us consider the given trigonometric identity $\tan \dfrac{u}{2}=\cos ecu-\cot u.$
We need to prove that the given identity is true. For that we need to show that the right-hand side of the equation can be derived from the left-hand side of the equation using some of the known identities.
Let us consider the left-hand side of the given identity, $\tan \dfrac{u}{2}.$
We know the identity $\tan \dfrac{x}{2}=\dfrac{1-\cos x}{\sin x}.$
Let us apply this identity on the left-hand side of our problem.
We will get $\tan \dfrac{u}{2}=\dfrac{1-\cos u}{\sin u}.$
Let us write this equation as $\tan \dfrac{u}{2}=\dfrac{1}{\sin u}-\dfrac{\cos u}{\sin u}.$
We are familiar with the basic trigonometric identity $\dfrac{1}{\sin x}=\cos ecx.$
Also, we know that the quotient we will get when we divide $\cos x$ by $\sin x,$ we will get Cotangent of $x.$ That is, $\dfrac{\cos x}{\sin x}=\cot x.$
Let us check if we will get the right-hand side of the given equation when we apply these identities in our equation.
From these identities, we will get $\dfrac{1}{\sin u}=\cos ecu$ and $\dfrac{\cos u}{\sin u}=\cot u.$
Let us apply the above identities in the equation we have derived using a known identity to get $\tan \dfrac{u}{2}=\cos ecu-\cot u.$
Therefore, the LHS is equal to the RHS. That is, LHS=RHS.
Hence, we have proved the given identity.

Note: In Mathematics, we can prove all the identities. Also, these identities can be used to prove other identities. Also, we have to be careful while doing calculations to avoid mistakes and errors.