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Question: How do you prove that \(\tan \left( {x + \left( {\dfrac{\pi }{2}} \right)} \right) = - \cot x\) ?...

How do you prove that tan⁡(x+(π2))=−cot⁡x\tan \left( {x + \left( {\dfrac{\pi }{2}} \right)} \right) = - \cot x ?

Explanation

Solution

By using the basic trigonometric identity given below we can prove the above expression that is tan⁡(x+(π2))=−cot⁡x\tan \left( {x + \left( {\dfrac{\pi }{2}} \right)} \right) = - \cot x . We can prove the given statement by using tan⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡bcos⁡acos⁡b−sin⁡asin⁡b\tan (a + b) = \dfrac{{\sin a\cos b + \cos a\sin b}}{{\cos a\cos b - \sin a\sin b}} . In order to solve and simplify the given expression we have to use the identity and express our given expression in the simplest form and thereby solve it.

Complete step-by-step solution:
To prove: tan⁡(x+(π2))=−cot⁡x\tan \left( {x + \left( {\dfrac{\pi }{2}} \right)} \right) = - \cot x
Proof:
Taking L.H.S of the above expression ,
tan⁡(x+(π2))\tan \left( {x + \left( {\dfrac{\pi }{2}} \right)} \right)
We already know that tan⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡bcos⁡acos⁡b−sin⁡asin⁡b\tan (a + b) = \dfrac{{\sin a\cos b + \cos a\sin b}}{{\cos a\cos b - \sin a\sin b}} ,
Therefore, by applying above identity we can write as ,
⇒tan⁡(x+(π2))=sin⁡xcos⁡π2+cos⁡xsin⁡π2cos⁡xcos⁡π2−sin⁡xsin⁡π2\Rightarrow \tan \left( {x + \left( {\dfrac{\pi }{2}} \right)} \right) = \dfrac{{\sin x\cos \dfrac{\pi }{2} + \cos x\sin \dfrac{\pi }{2}}}{{\cos x\cos \dfrac{\pi }{2} - \sin x\sin \dfrac{\pi }{2}}}
We know that cos⁡π2=0\cos \dfrac{\pi }{2} = 0 and sin⁡π2=1\sin \dfrac{\pi }{2} = 1
⇒sin⁡x(0)+cos⁡x(1)cos⁡x(0)−sin⁡x(1)\Rightarrow \dfrac{{\sin x (0) + \cos x (1) }}{{\cos x (0) - \sin x (1) }}
⇒cos⁡x−sin⁡x\Rightarrow \dfrac{{\cos x}}{{ - \sin x}}
We know that cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}}
Therefore , we will get ,
⇒−cot⁡x\Rightarrow - \cot x
It is equal to R.H.S. of the given problem ,
Since L.H.S. = R.H.S.
Hence proved.

Hence it is proved that L.H.S = R.H.S

Note: Some other equations needed for solving these types of problem are:
tan⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡bcos⁡acos⁡b−sin⁡asin⁡b\tan (a + b) = \dfrac{{\sin a\cos b + \cos a\sin b}}{{\cos a\cos b - \sin a\sin b}} ,
cos⁡π2=0\cos \dfrac{\pi }{2} = 0 , sin⁡π2=1\sin \dfrac{\pi }{2} = 1 ,and cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{{\cos x}}{{\sin x}} .
Range of cosine and sine is [−1,1]\left[ { - 1,1} \right] ,
Range of tangents is minus infinity to infinity.

In order to solve and simplify the given expression we have to use the identity and express our given expression in the simplest form and thereby solve it. Also, while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.