In Δ A B C if θ is any angle, then b cos (C+θ)+c cos (B-θ)=

Question

In Δ A B C if θ is any angle, then b cos (C+θ)+c cos (B-θ)= (A) a cot θ (B) a cos θ (C) a tan θ (D) a sin θ

Tardigrade Admin · Accepted Answer

Consider, b cos (C+θ)+c cos (B-θ) = b( cos C cos θ- sin C sin θ)+c ( cos B cos θ+ sin B sin θ) = cos θ(b cos C+c cos B)- sin θ (b sin C-c sin B) = a cos θ- sin θ .0 [∵ By projection formula and by sine rule (b/ sin B) =(c/ sin C) ⇒ b sin C-c sin B=0] =a cos θ- sin θ ⋅ 0=a cos θ

Q. In $Δ A BC$ if $θ$ is any angle, then $b cos (C + θ) + c cos (B - θ) =$

$a cot θ$

$a cos θ$

$a tan θ$

$a sin θ$

Q. In ΔABC if θ is any angle, then bcos(C+θ)+ccos(B−θ)=

acotθ

acosθ

atanθ

asinθ

Consider, bcos(C+θ)+ccos(B−θ) =b(cosCcosθ−sinCsinθ)+c (cosBcosθ+sinBsinθ) =cosθ(bcosC+ccosB)−sinθ (bsinC−csinB) =acosθ−sinθ.0 [∵ By projection formula and by sine rule sinBb​=sinCc​ ⇒bsinC−csinB=0] =acosθ−sinθ⋅0=acosθ

Q. In $Δ A BC$ if $θ$ is any angle, then $b cos (C + θ) + c cos (B - θ) =$

$a cot θ$

$a cos θ$

$a tan θ$

$a sin θ$