Value of - sinB-Ccos B cos C - sinC-Acos C cos A -sinA-Bcos A cos B

The correct option is B 0 sin(B C)cos B cos C+sin(C A)cos C cos A +sin(A B)cos A cos B =sin B cos C cos B sin Ccos B cos C+sin C cos A cos C sin Acos ...

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Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

Value of : s...

Question

Value of :

$\frac{sin (B - C)}{cos B cos C} + \frac{sin (C - A)}{cos C cos A} + \frac{sin (A - B)}{cos A cos B}$

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Solution

The correct option is B 0
$\frac{sin (B - C)}{cos B cos C} + \frac{sin (C - A)}{cos C cos A} + \frac{sin (A - B)}{cos A cos B}$

$= \frac{sin B cos C - cos B sin C}{cos B cos C} + \frac{sin C cos A - cos C sin A}{cos C cos A} + \frac{sin A cos B - cos A sin B}{cos A cos B}$

$= tan B - tan C + tan C - tan A + tan A - tan B = 0$

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Similar questions

Q. If

A, B, C \neq \frac{(2 n + 1) π}{2}

, then the numerical value of

\frac{sin (B - C)}{cos B cos C} + \frac{sin (C - A)}{cos C cos A} + \frac{sin (A - B)}{cos A cos B}

Q. If

A, B, C \neq \frac{(2 n + 1) π}{2}

, then the numerical value of

\frac{sin (B - C)}{cos B cos C} + \frac{sin (C - A)}{cos C cos A} + \frac{sin (A - B)}{cos A cos B}

Q. Prove that:
(i)

\frac{s i n (A + B) + s i n (A - B)}{c o s (A + B) + c o s (A - B)} = t a n A

(ii)

\frac{s i n (A - B)}{c o s A c o s B} + \frac{s i n (B - C)}{c o s B c o s C} + \frac{s i n (C - A)}{c o s C c o s A} = 0

(iii)

\frac{s i n (A - B)}{s i n A s i n B} + \frac{s i n (B - C)}{s i n B s i n C} + \frac{s i n (C - A)}{s i n C s i n A} = 0

(iv)

s i n^{2} B = s i n^{2} A + s i n^{2} (A - B) - 2 s i n A c o s B s i n (A - B)

(v)

c o s^{2} + c o s^{2} B - 2 c o s A c o s B c o s (A + B) = s i n^{2} (A + B)

(vi)

\frac{t a n (A + B)}{c o t (A - B)} = \frac{t a n^{2} A - t a n^{2} B}{1 - t a n^{2} A t a n^{2} B}

Q. Find the value of :

\frac{a^{2} sin (B - C)}{sin B + sin C} + \frac{b^{2} sin (C - A)}{sin C + sin A} + \frac{c^{2} sin (A - B)}{sin A + sin B}

Q. In a

Δ A B C

{cos}^{2} A + {cos}^{2} B + {cos}^{2} C = cos A cos B + cos B cos C + cos C cos A

, then the triangle is

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Value of :sin(B−C)cosBcosC+sin(C−A)cosCcosA+sin(A−B)cosAcosB

The correct option is B 0sin(B−C)cosBcosC+sin(C−A)cosCcosA+sin(A−B)cosAcosB=sinBcosC−cosBsinCcosBcosC+sinCcosA−cosCsinAcosCcosA+sinAcosB−cosAsinBcosAcosB=tanB−tanC+tanC−tanA+tanA−tanB=0

Value of :

$\frac{sin (B - C)}{cos B cos C} + \frac{sin (C - A)}{cos C cos A} + \frac{sin (A - B)}{cos A cos B}$