How do you write the equation as the sine, cosine or tangent of the angle given
\[\dfrac{\tan {{25}^{\circ }}+\tan {{10}^{\circ }}}{1-\tan {{25}^{\circ }}\tan {{10}^{\circ }}}\]

Hint: In this problem, we have to write the equation as the sine, cosine, or tangent of the angle given \[\dfrac{\tan {{25}^{\circ }}+\tan {{10}^{\circ }}}{1-\tan {{25}^{\circ }}\tan {{10}^{\circ }}}\]. To solve these types of problems, we should know some trigonometric formulas and identities. We have a trigonometric identity to simplify the given expression and to find the answer.

Complete step by step answer:
We know that the given trigonometric expression to be solved is,
\[\dfrac{\tan {{25}^{\circ }}+\tan {{10}^{\circ }}}{1-\tan {{25}^{\circ }}\tan {{10}^{\circ }}}\]……. (1)
We know that the trigonometric formula, which is found using sin and cosine angle addition formulas which is used to solve this problem is,
\[\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\tan \left( x+y \right)\]….. (2)
Now we can use the above formula to solve this question.
We can apply the above formula in the trigonometric expression (1).
We can take the left-side of the given expression and left-hand side of the trigonometric formula. By comparing both, we can say that
x = \[{{25}^{\circ }}\] and y = \[{{10}^{\circ }}\].
We can substitute the above values in the trigonometric identity (2), we get
\[\Rightarrow \dfrac{\tan {{25}^{\circ }}+\tan {{10}^{\circ }}}{1-\tan {{25}^{\circ }}\tan {{10}^{\circ }}}=\tan \left( {{25}^{\circ }}+{{10}^{\circ }} \right)\]
\[\Rightarrow \tan {{35}^{\circ }}\]
Therefore, the value of \[\dfrac{\tan {{25}^{\circ }}+\tan {{10}^{\circ }}}{1-\tan {{25}^{\circ }}\tan {{10}^{\circ }}}\] is \[\tan {{35}^{\circ }}\].

Note:
Students make mistakes while adding the degree values at the end of the problem, which should be concentrated. We should know that to solve these types of problems, we have to know some trigonometric formulas and identities. We should also know the degree values for sine, cosine, or tangent to find the exact value if needed.