# Consider the angle sum formulas: sin(A+B)=sinA cosB+cosA sinB cos(A+B)=cosA cosB−sinA sinB show that tan(90^0+θ)=− 1/tanθ. and use this identity to prove that lines y = mx and y = m_1x are perpendicular iff mm_1=−1?

Jan 8, 2017

Given

$\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B \ldots . \left[1\right]$

$\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B \ldots . \left[2\right]$

So $\tan \left(A + B\right) = \sin \frac{A + B}{\cos} \left(A + B\right)$

$= \frac{\frac{\sin A \cos B}{\cos A \cos B} + \frac{\cos A \sin B}{\cos A \cos B}}{\frac{\cos A \cos B}{\cos A \cos B} - \frac{\sin A \sin B}{\cos A \cos B}}$

$\implies \tan \left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \ldots . . \left[3\right]$

$\implies \tan \left(A + B\right) = \frac{\tan \frac{A}{\tan} A + \tan \frac{B}{\tan} A}{\frac{1}{\tan} A - \tan B}$

$\implies \tan \left(A + B\right) = \frac{1 + \tan B \cdot \cot A}{\cot A - \tan B} \ldots . . \left[4\right]$

This is an identity so it is valid for real values of $A \mathmr{and} B$.So putting $A = {90}^{\circ} \mathmr{and} B = \theta$ in [4] we get

$\implies \tan \left({90}^{\circ} + \theta\right) = \frac{1 + \tan \theta \cdot \cot 90}{\cot 90 - \tan \theta}$

$\implies \tan \left({90}^{\circ} + \theta\right) = \frac{1 + \tan \theta \cdot 0}{0 - \tan \theta} = - \frac{1}{\tan} \theta$

$\implies \tan \left({90}^{\circ} + \theta\right) = - \frac{1}{\tan} \theta \ldots . \left[5\right]$

Now if $y = m x$ straight line makes angle $\theta$ with positive direction of x-axis then $m = \theta$.Again if the straight line having equation $y = {m}_{1} x$ makes an angle ${90}^{\circ} + \theta$ with the positive direction of x-axis,then it will be perpendicular to the first straight line and its slope m_1=tan(90^@+theta#

So by relation [5] we get

${m}_{1} = - \frac{1}{m}$

$m {m}_{1} = - 1$, for mutually perpendicular straight lines