Question

How do you prove $\tan \left( {{90}^{\circ }}+a \right)=-\cot \left( a \right)$ ?

Answer 1

Hint: We are asked to prove that $\tan \left( {{90}^{\circ }}+a \right)=-\cot \left( a \right)$ . To do so, we will learn how $\tan \theta $ is connected to other ratio, we will use that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ , then we will learn how sin and cos expand when they have sum of two terms.
We will use $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ and $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ .
Lastly we will need the knowledge that $\sin {{90}^{\circ }}=1$ and $\cos {{90}^{\circ }}=0$ .

Complete step by step solution:
We are given to prove that $\tan \left( {{90}^{\circ }}+a \right)=-\cot \left( a \right)$ .
We can see that the left side has $\tan \left( {{90}^{\circ }}+a \right)$ .
It has more terms, so it will be helpful for us to start from there and return to the right hand side that will achieve $-\cot \left( a \right)$ .
To do so, we will start by considering the left hand side.
So, we have $\tan \left( {{90}^{\circ }}+a \right)$ .
Now as we know that tan is nothing but the ratio of sin and cos i.e. $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ .
So, using this on ${{90}^{\circ }}+a$ , we get –
$\Rightarrow \tan \left( {{90}^{\circ }}+a \right)=\dfrac{\sin \left( {{90}^{\circ }}+a \right)}{\cos \left( {{90}^{\circ }}+a \right)}$ ……………………………………. (1)
Now, to solve it further we will expand our sin and cos function.
As we know that $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ .
So, using $A={{90}^{\circ }}$ and $B=\theta $ , we get –
$\Rightarrow \sin \left( {{90}^{\circ }}+a \right)=\sin {{90}^{\circ }}\operatorname{cosa}+\cos {{90}^{\circ }}\sin a$
As $\sin {{90}^{\circ }}=1$ and $\cos {{90}^{\circ }}=0$
So, we get –
$\begin{align}
  & \Rightarrow \cos a+0 \\
 & =\cos a \\
\end{align}$
So, $\sin \left( {{90}^{\circ }}+a \right)=\cos a$ ……………………………… (2)
Now as we know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ .
So using ${{90}^{\circ }}$ as A and $\theta $ as B, we get –
$\Rightarrow \cos \left( {{90}^{\circ }}+a \right)=\cos {{90}^{\circ }}\cos a-\sin {{90}^{\circ }}\sin a$
As $\sin {{90}^{\circ }}=1$ and $\cos {{90}^{\circ }}=0$ , so we get –
 $\begin{align}
  & =0-\sin a \\
 & =-\sin a \\
\end{align}$
So, $\cos \left( {{90}^{\circ }}+a \right)=-\sin a$ ………………………… (3)
So, using equation (2) and (3) in equation (1), we get –
$\Rightarrow \tan \left( {{90}^{\circ }}+a \right)=\dfrac{\sin \left( {{90}^{\circ }}+a \right)}{\cos \left( {{90}^{\circ }}+a \right)}$
As $\sin \left( {{90}^{\circ }}+a \right)=\cos a$ and $\cos \left( {{90}^{\circ }}+a \right)=-\sin a$
So, $=\dfrac{\cos a}{-\sin a}$
By simplifying, we get –
$=-\dfrac{\cos a}{\sin a}$
As we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ .
So, we get –
$\Rightarrow \dfrac{\cos a}{\sin a}=\cot a$

Hence we get –
$\tan \left( {{90}^{\circ }}+a \right)=-\cot a$.

Note: Just like this there are other ratio which are related to one another, we can use similar approach to solve such relation are –
1, $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $
2, $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta $
3, $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $
4, $\sec \left( {{90}^{\circ }}-\theta \right)=\cos ec\theta $
5, $\cos \left( {{180}^{\circ }}+x \right)=-\cos x$
6, $\sin \left( {{180}^{\circ }}+x \right)=-\sin x$
We can use the same way to verify these identity, remember $\sin \left( {{180}^{\circ }} \right)=0$ and $\cos \left( {{180}^{\circ }} \right)=-1$ .