First slide

Vectors

Question

Vectors $\vec{A} a n d \vec{B} include an angle θ between them . If (\vec{A} + \vec{B}) and (\vec{A} - \vec{B}) respectively subtend angles α and β with A, then (tanα + tanβ) is$

Moderate

A

$\frac{(AB sinθ)}{(A^{2} + B^{2} \cos^{2} θ)}$
B

$\frac{(2 AB sinθ)}{(A^{2} - B^{2} \cos^{2} θ)}$
C

$\frac{(A^{2} \sin^{2} θ)}{(A^{2} + B^{2} \cos^{2} θ)}$
D

$\frac{(B^{2} \sin^{2} θ)}{(A^{2} - B^{2} \cos^{2} θ)}$

Solution

$tanα = \frac{Bsinθ}{A + Bcosθ} - - - - - - (i)$
$Where α is the angle made by the vector (\vec{A} + \vec{B}) with \vec{A}$
Similarly, $tanβ = \frac{Bsinθ}{A - Bcosθ}$ -------(ii)
Where $β$ is the angle made by the vector $(\vec{A} - \vec{B}) with \vec{A}$
Note that the angle between $\vec{A} and (- \vec{B})$ is $(180^{0}) - θ$
Adding (i) and (ii), we get
$tanα + tanβ = \frac{Bsinθ}{A + Bcosθ} + \frac{Bsinθ}{A - Bcosθ}$
$= \frac{(AB sinθ - B^{2} sinθcosθ + ABsinθ + B^{2} sinθcosθ}{(A + Bcosθ) (A - Bcosθ)}$
= $\frac{2 ABsinθ}{(A^{2} - B^{2} \cos^{2} θ)}$

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