#### Need solution for RD Sharma maths class 12 chapter 5 Determinants exercise multiple choise question 35

Correct option (a)

Hint:

A, B and C are angle of a $\Delta$, then ∠A+∠B+∠C=180  and solve determinant.

Given:

Given that,

A, B and C are angle of a triangle

We have to find the value of

$|-1 \cos \cos C \cos \cos B \cos \cos C-1 \cos \cos A \cos \cos B \cos \cos A-1|$

Solution:

We have

$|-1 \cos \cos C \cos \cos B \cos \cos C-1 \cos \cos A \cos \cos B \cos \cos A-1|$

Expanding along R1, we have

\begin{aligned} &=-1\left(1-\cos ^{2} A\right)-\cos C(-\cos C-\cos A \cos B)+\cos B(\cos A \cos C+\cos B) \\ &=-1+\cos ^{2} A+\cos ^{2} C+\cos A \cos B \cos C+\cos A \cos B \cos C+\cos ^{2} B \\ &=\cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &=\left(\frac{1+\cos 2 A}{2}\right)+\left(\frac{1+\cos 2 B}{2}\right)+\cos ^{2} C+2 \cos A \cos B \cos C-1 \quad\left[\because A=\frac{1+\cos \cos 2 A}{2}\right] \\ &=\frac{2+\cos 2 A+\cos 2 B}{2}+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &=\frac{2+2 \cos \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)}{2}+\cos ^{2} C+2 \cos A \cos B \cos C-1 \end{aligned}

\begin{aligned} &=1+\cos (A+B) \cos (A-B)+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &=1+\cos (180-C) \cos (A-B)+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &{[\angle A+\angle B+\angle C=180]} \\ &=1-\cos C \cos (A-B)+\cos ^{2} C+2 \cos A \cos B \cos C-1 \\ &=\cos \cos C[-\cos \cos (A-B)+\cos \cos C]+2 \cos \cos A \cos \cos B \cos \cos C \\ &=\cos \cos C[-\cos \cos (A-B)-\cos \cos (A+B)]+2 \cos \cos A \cos \cos B \cos \cos C \\ &{[\because \cos \cos C=\cos \cos (180-(A+B))=-\cos \cos (A+B)]} \\ &=-\cos \cos C\left[2 \cos \cos \left(\frac{A-B+A+B}{2}\right) \cos \cos \left(\frac{A-B-A-B}{2}\right)\right]+2 \cos \cos A \\ &\cos \cos B \cos \cos C \\ &=-2 \cos C \cos A \cos B+2 \cos A \cos B \cos C \\ &=0 \end{aligned}

Hence 0 is the required answer.