How do you verify \[\tan \left( \theta +\dfrac{\pi }{2} \right)=-\cot \theta \]?

Hint: To prove a statement, we must show that either the left-hand side or right-hand side can be expressed as the other side. To prove this statement, we must know the trigonometric expansion formula of \[\tan (A+B)\]. \[\tan (A+B)\]is expanded as \[\dfrac{\tan A+\tan B}{1-\tan A\tan B}\]. We will use this expansion formula to solve this question.

Complete step-by-step answer:
We are asked to prove the statement \[\tan \left( \theta +\dfrac{\pi }{2} \right)=-\cot \theta \]. The LHS of the above statement is \[\tan \left( \theta +\dfrac{\pi }{2} \right)\], and the right-hand side of the statement is \[-\cot \theta \]. Let’s take the left-hand side of the expression. We can use the expansion formula of \[\tan (A+B)\], here we have \[A=\theta \And B=\dfrac{\pi }{2}\].
We know the expansion formula for \[\tan (A+B)\] is \[\dfrac{\tan A+\tan B}{1-\tan A\tan B}\], substituting the values of A, and B. we get
 \[\Rightarrow \dfrac{\tan \theta +\tan \dfrac{\pi }{2}}{1-\tan \theta \tan \dfrac{\pi }{2}}\]
Dividing the numerator and denominator by \[\tan \dfrac{\pi }{2}\], we get
\[\Rightarrow \dfrac{\dfrac{\tan \theta +\tan \dfrac{\pi }{2}}{\tan \dfrac{\pi }{2}}}{\dfrac{1-\tan \theta \tan \dfrac{\pi }{2}}{\tan \dfrac{\pi }{2}}}=\dfrac{\dfrac{\tan \theta }{\tan \dfrac{\pi }{2}}+1}{\dfrac{1}{\tan \dfrac{\pi }{2}}-\tan \theta }\]
We know that \[\dfrac{1}{\tan a}=\cot a\], using this in the above expression, it can be written as
\[\Rightarrow \dfrac{\tan \theta \cot \dfrac{\pi }{2}+1}{\cot \dfrac{\pi }{2}-\tan \theta }\]
We know \[\cot \dfrac{\pi }{2}=0\], substituting its value in the above expression, we get
\[\Rightarrow \dfrac{0+1}{0-\tan \theta }=\dfrac{1}{-\tan \theta }\]
Again, using the property \[\dfrac{1}{\tan a}=\cot a\], the above expression can be written as
\[\Rightarrow -\cot \theta \]
Hence, as the left-hand side of the given statement can be expressed as the right-hand side. This statement is proved.

Note: The given statement is one of the important trigonometric properties, and so it should be remembered. As it will be useful in solving other trigonometric questions including evaluation of expression or proofs. We can also prove the properties \[\sin \left( \theta +\dfrac{\pi }{2} \right)=\cos \theta \], and \[\text{cosec}\left( \theta +\dfrac{\pi }{2} \right)=\sec \theta \] by a similar method.