Vectors vecA and vecB include an angle θ between them. If ( vecA+ vecB) and ( vecA- vecB) respectively subtend angles α and β with vecA, then (tan α+tan β) is

Question

Vectors vecA and vecB include an angle θ between them. If ( vecA+ vecB) and ( vecA- vecB) respectively subtend angles α and β with vecA, then (tan α+tan β) is (A) ((AB sin θ)/(A2+B2 cos2 θ)) (B) ((2AB sin θ)/(A2-B2 cos2 θ)) (C) ((A2 sin2 θ)/(A2+B2 cos2 θ )) (D) ((B2 sin2 θ)/(A2-B2 cos2 θ ))

Tardigrade Admin · Accepted Answer

tan α=(B sin θ/A+B cos θ) ...(i) where α is the angle made by the vector ( vecA+ vecB) with vecA. Similarly, tan β=(B sin θ/A-B cos θ) ...(ii) where β is the angle made by the vector ( vecA- vecB) with vecA. Note that the angle between vecA. and (- vecB) is (180°-θ). Adding (i) and (ii), we get tan α+tan β=(B sin θ/A+B cos θ)+(B sin θ/A-B cos θ) =(AB sin θ-B2 sin θ cos θ+AB sin θ+B2 sin θ cos θ/(A+B cos θ)(A-B cos θ)) =(2AB sin θ/(A2-B2 cos2 θ))

Q. Vectors $A$ and $B$ include an angle $θ$ between them. If $(A + B)$ and $(A - B)$ respectively subtend angles $α$ and $β$ with $A$ , then $(t an α + t an β)$ is

$\frac{( A B s in θ )}{( A ^{2} + B ^{2} co s ^{2} θ )}$

$\frac{( 2 A B s in θ )}{( A ^{2} - B ^{2} co s ^{2} θ )}$

$\frac{( A ^{2} s i n ^{2} θ )}{( A ^{2} + B ^{2} co s ^{2} θ )}$

$\frac{( B ^{2} s i n ^{2} θ )}{( A ^{2} - B ^{2} co s ^{2} θ )}$

Q. Vectors A and B include an angle θ between them. If (A+B) and (A−B) respectively subtend angles α and β with A, then (tanα+tanβ) is

(A2+B2cos2θ)(ABsinθ)​

(A2−B2cos2θ)(2ABsinθ)​

(A2+B2cos2θ)(A2sin2θ)​

(A2−B2cos2θ)(B2sin2θ)​

Q. Vectors $A$ and $B$ include an angle $θ$ between them. If $(A + B)$ and $(A - B)$ respectively subtend angles $α$ and $β$ with $A$ , then $(t an α + t an β)$ is

$\frac{( A B s in θ )}{( A ^{2} + B ^{2} co s ^{2} θ )}$

$\frac{( 2 A B s in θ )}{( A ^{2} - B ^{2} co s ^{2} θ )}$

$\frac{( A ^{2} s i n ^{2} θ )}{( A ^{2} + B ^{2} co s ^{2} θ )}$

$\frac{( B ^{2} s i n ^{2} θ )}{( A ^{2} - B ^{2} co s ^{2} θ )}$