Question

How do you prove $\tan \left(90°+x\right)=-cot\left(x\right)$?

Answer 1

Proof of given relation:

$\tan \left(90°+x\right)=-cot\left(x\right)$

Here we have $LHS=\tan \left(90°+x\right)$

Therefore,

$\tan \left(90°+x\right)=\frac{\sin \left(90°+x\right)}{\cos \left(90°+x\right)}$ $\left[\because \mathrm{tanx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}\right]$

$\begin{array}{l}=\frac{sin\left(x\right)cos(90°)+cos\left(x\right)sin(90°)}{cos\left(x\right)cos(90°)–sin\left(x\right)sin(90°)}\end{array}$ $$\begin{gathered} \left[\because \sin \left(a+b\right)=\mathrm{sina}\mathrm{cosb}+\mathrm{cosa}\mathrm{sinb}\right] \\ \left[\because \cos \left(a+b\right)=\mathrm{cosa}\mathrm{cosb}-\mathrm{sina}\mathrm{sinb}\right] \end{gathered}$$

$=\frac{\cos x+0}{0-\sin x}$ $\left[\because \sin 90°=1;\cos 90°=0\right]$

$=-\frac{\cos x}{\sin x}$

$=-cotx$ $\left[\because cotx=\frac{cosx}{sinx}\right]$

$=RHS$

Hence, $\mathbf{tan}\left(90°+x\right)\mathbf{=}\mathbf{-}\mathbf{cot}\left(x\right)$ is proved.