If $A + B + C = 180^{\circ},$ then secA(cosBcosC -sinBsinC) is equal to

-1

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None of these

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Solution

The correct option is A -1
secA(cosBcosC -sinBsinC) = $\frac{c o s B c o s (π - (A + B)) - s i n B s i n (π - (A + B))}{c o s A}$
We know that, $c o s (π - θ) = - c o s θ$ and $s i n (π - θ) = s i n θ$
$∴ s e c A (c o s B c o s C - s i n B s i n C) = \frac{c o s B c o s (A + B) - s i n B s i n (A + B)}{c o s A}$
Now, using the identities cos(A+B)=cosAcosB-sinAsinB and sin(A+B)=sinAcosB+cosAsinB, we get
secA(cosBcosC-sinBsinC) = $\frac{- c o s A c o s B^{2} + c o s B s i n A s i n B - s i n B s i n A c o s B - s i n^{2} B c o s A}{c o s A}$
$\Rightarrow s e c A (c o s B c o s C - s i n B s i n C) = - \frac{c o s A (c o s^{2} B + s i n^{2} B)}{c o s A}$
$\Rightarrow s e c A (c o s B c o s C - s i n B s i n C) = \frac{- c o s A}{c o s A} = - 1$

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Similar questions

|\begin{matrix} - 1 & \cos C & \cos B \\ \cos C & - 1 & \cos A \\ \cos B & \cos A & - 1 \end{matrix}|

π

s e c A (c o s B c o s C - s i n B s i n C)

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