If A-B-C- then sinA-B-C sin B cos C -sinB 0 tan A cosA-B -tanA 0 equals

The correct option is A 0 Δ =sin( A+B+C ) #160; #160; amp; sin B #160; amp; cos C #160; sin B #160; amp; 0 amp; tan A #160; ...

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Determinant

If A+B+C=π,...

Question

If $A + B + C = π$ , then $∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n (A + B + C) & s i n B & c o s C - s i n B & 0 & t a n A c o s (A + B) & - t a n A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$ equals

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2 sin B tan A cos C

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none of these

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Solution

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A + B + C = π

∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n (A + B + C) & s i n B & c o s C - s i n B & 0 & t a n A c o s (A + B) & - t a n A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ =

A + B + C = π

∣ ∣ ∣ ∣ ∣ \begin{matrix} sin (A + B + C) & sin B & cos C - sin B & 0 & tan A cos (A + B) & - tan A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

A + B + C = π

∣ ∣ ∣ ∣ ∣ \begin{matrix} sin (A + B + C) & sin B & cos C sin B & 0 & tan A cos (A + B) & - tan A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ =

A + B + C = π

∣ ∣ ∣ ∣ ∣ \begin{matrix} tan (A + B + C) & tan B & tan C tan (A + C) & 0 & tan A tan (A + B) & - tan A & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

0 \leq x \leq 1

(a, b) =

(0, π)

(0, π / 4)

(- π / 4, π / 4)

(π / 4, π / 2)

a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

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