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Question: Vectors \(\overrightarrow{A}\)and \(\overrightarrow{B}\)include an angle \(\theta\)between them. If ...

Vectors A→\overrightarrow{A}and B→\overrightarrow{B}include an angle θ\thetabetween them. If (A→+B→)(\overrightarrow{A} + \overrightarrow{B}) and (A→−B→)(\overrightarrow{A} - \overrightarrow{B}) respectively subtend angles α\alphaand β\betawith A→\overrightarrow{A}, then (tan⁡α+tan⁡β)(\tan\alpha + \tan\beta) is

A

(ABsin⁡θ)(A2+B2cos⁡2θ)\frac{(AB\sin\theta)}{(A^{2} + B^{2}\cos^{2}\theta)}

B

(2ABsin⁡θ)(A2−B2cos⁡2θ)\frac{(2AB\sin\theta)}{(A^{2} - B^{2}\cos^{2}\theta)}

C

(A2sin⁡2θ)(A2+B2cos⁡2θ)\frac{(A^{2}\sin^{2}\theta)}{(A^{2} + B^{2}\cos^{2}\theta)}

D

(B2sin⁡2θ)(A2−B2cos⁡2θ)\frac{(B^{2}\sin^{2}\theta)}{(A^{2} - B^{2}\cos^{2}\theta)}

Answer

(2ABsin⁡θ)(A2−B2cos⁡2θ)\frac{(2AB\sin\theta)}{(A^{2} - B^{2}\cos^{2}\theta)}

Explanation

Solution

tan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} ……(i)

Where α\alphais the angle made by the vector

(A→+B→)(\overset{\rightarrow}{A} + \overset{\rightarrow}{B}) with A→\overset{\rightarrow}{A}.

Similarly, tan⁡β=Bsin⁡θA−Bcos⁡θ\tan\beta = \frac{B\sin\theta}{A - B\cos\theta} …... (ii)

Where β\betais the angle made by the vector (A→−B→)(\overset{\rightarrow}{A} - \overset{\rightarrow}{B}) with A→\overset{\rightarrow}{A}.

Note that the angle between A→\overset{\rightarrow}{A} and (−B→- \overset{\rightarrow}{B})is

(1800−θ).(180^{0} - \theta). Adding (i) and (ii), we get

tan⁡α+tan⁡β=Bsin⁡θA+Bcos⁡θ+Bsin⁡θA−Bcos⁡θ\tan\alpha + \tan\beta = \frac{\mathbf{B}\sin\theta}{A + B\cos\theta} + \frac{B\sin\theta}{A - B\cos\theta}

=ABsin⁡θ−B2sin⁡θcos⁡θ+ABsin⁡θ+B2sin⁡θcos⁡θ(A+Bcos⁡θ)(A−Bcos⁡θ)=2ABsin⁡θ(A2−B2cos⁡2θ)= \frac{AB\sin\theta - B^{2}\sin\theta\cos\theta + AB\sin\theta + B^{2}\sin\theta\cos\theta}{(A + B\cos\theta)(A - B\cos\theta)} = \frac{2AB\sin\theta}{(A^{2} - B^{2}\cos^{2}\theta)}