C-b cos A-b-c cos A-cos B-cos C

LHS =c b cos A/b c cos A =K sin C K sin B cos A/K sin B K sin C cos A =sin(π (A B ) ) sin B cos A/sin(π (A+C ) ) sin C cos A =sin(A+B ) sin B cos A/sin ...

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Byju's Answer

Standard XII

Mathematics

Sine Rule

c-b cos A/b-c...

Question

$\frac{c - b c o s A}{b - c c o s A} = \frac{c o s B}{c o s C}$

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Solution

$L H S = \frac{c - b c o s A}{b - c c o s A} = \frac{K s i n C - K s i n B c o s A}{K s i n B - K s i n C c o s A} = \frac{s i n (π - (A - B)) - s i n B c o s A}{s i n (π - (A + C)) - s i n C c o s A} = \frac{s i n (A + B) - s i n B c o s A}{s i n (A + C) - s i n C c o s A} = \frac{s i n A c o s B + c o s A s i n B - s i n B c o s A}{s i n A c o s C + c o s A s i n C - s i n C c o s A} = \frac{s i n A c o s B}{s i n A c o s C} = \frac{c o s B}{c o s C} = R H S$

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Similar questions

Prove:

\frac{c - b cos A}{b - c cos A} = \frac{cos B}{cos C}

Q. Prove that,

\frac{c - b c o s A}{b - c c o s A} = \frac{c o s B}{c o s C}

Q. Prove that

$\frac{c o s A}{c c o s B + b c o s C} + \frac{c o s B}{a c o s C + c c o s A} + \frac{c o s C}{a c o s B + b c o s A} = \frac{a^{2} + b^{2} + c^{2}}{2 a b c}$

Q. Solve:

\frac{c - b c o s A}{b - c c o s A} = \frac{c o s A}{c o s C}

Q. In any

△ A B C

prove the following:

(i) a s i n A - b s i n B = c s i n (A - B)

(i i) \frac{c - b c o s A}{b - c c o s A} = \frac{c o s B}{c o s C}

(i i i) a^{2} s i n (B - C) = (b^{2} - c^{2}) s i n A

(i v) \frac{(a - b)}{(a + b)} = \frac{tan (\frac{A - B}{2})}{t a n (\frac{A + B}{2})}

(v) a c o s B - b c c o s A = (a^{2} - b 2)

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c−b cos Ab−c cos A=cos Bcos C

LHS=c−b cos Ab−c cos A=K sin C−K sin B cos AK sin B−K sin C cos A=sin(π−(A−B))−sin B cos Asin(π−(A+C))−sin C cos A=sin(A+B)−sin B cos Asin(A+C)−sin C cos A=sin A cos B+cos A sin B−sin B cos Asin A cos C+cos A sin C−sin C cos A=sin A cos Bsin A cos C=cos Bcos C=RHS

$\frac{c - b c o s A}{b - c c o s A} = \frac{c o s B}{c o s C}$