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

How do you verify tan(θ+π2)=cotθ\tan \left( \theta +\dfrac{\pi }{2} \right)=-\cot \theta ?

Explanation

Solution

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). tan(A+B)\tan (A+B)is expanded as tanA+tanB1tanAtanB\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(θ+π2)=cotθ\tan \left( \theta +\dfrac{\pi }{2} \right)=-\cot \theta . The LHS of the above statement is tan(θ+π2)\tan \left( \theta +\dfrac{\pi }{2} \right), and the right-hand side of the statement is cotθ-\cot \theta . Let’s take the left-hand side of the expression. We can use the expansion formula of tan(A+B)\tan (A+B), here we have A=θ&B=π2A=\theta \And B=\dfrac{\pi }{2}.
We know the expansion formula for tan(A+B)\tan (A+B) is tanA+tanB1tanAtanB\dfrac{\tan A+\tan B}{1-\tan A\tan B}, substituting the values of A, and B. we get
tanθ+tanπ21tanθtanπ2\Rightarrow \dfrac{\tan \theta +\tan \dfrac{\pi }{2}}{1-\tan \theta \tan \dfrac{\pi }{2}}
Dividing the numerator and denominator by tanπ2\tan \dfrac{\pi }{2}, we get
tanθ+tanπ2tanπ21tanθtanπ2tanπ2=tanθtanπ2+11tanπ2tanθ\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 1tana=cota\dfrac{1}{\tan a}=\cot a, using this in the above expression, it can be written as
tanθcotπ2+1cotπ2tanθ\Rightarrow \dfrac{\tan \theta \cot \dfrac{\pi }{2}+1}{\cot \dfrac{\pi }{2}-\tan \theta }
We know cotπ2=0\cot \dfrac{\pi }{2}=0, substituting its value in the above expression, we get
0+10tanθ=1tanθ\Rightarrow \dfrac{0+1}{0-\tan \theta }=\dfrac{1}{-\tan \theta }
Again, using the property 1tana=cota\dfrac{1}{\tan a}=\cot a, the above expression can be written as
cotθ\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(θ+π2)=cosθ\sin \left( \theta +\dfrac{\pi }{2} \right)=\cos \theta , and cosec(θ+π2)=secθ\text{cosec}\left( \theta +\dfrac{\pi }{2} \right)=\sec \theta by a similar method.