Question

Assertion :Let $a,b,c$ be the sides of a $\Delta ABC$ opposite angles $A,B,C$ respectively
Let $S={\sum }_{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)b^{n-r}{c}^r\cos \{rB-(n-r\left)C\right\}$
and $T={\sum }_{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)b^{n-r}{c}^r\sin \{rB-(n-r\left)C\right\}$
$S=a^n$ Reason: $T=a^n$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

$S+iT=\sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)b^{n-r}{c}^r{e}^{i[rB-(n-r\left)C\right]}$
$=\sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right){\left({be}^{-iC}\right)}^{n-r}{\left(c{e}^{iB}\right)}^r$
$={\left[\right(b\cos C+c\cos B)+i(-b\sin C+c\sin B\left)\right]}^n$
But $b\cos C+c\cos B=a$ and $\frac{b}{\sin B}=\frac{c}{\sin C}$
Thus $S+iT=a^n$ $\Rightarrow S=a^n$ and $T=0$