Sin B-C -cos Bcos C-sin C-A -cos Ccos A-sin A-B -cos Acos B-

nbsp;The first term of L.H.S. IS sin ( B C )/cos Bcos C=sin Bcos C cos Bsin C/cos Bcos C=sin Bcos C/cos Bcos C cos Bsin C/cos Bcos C=tan B tan C Simila ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

sin B-C /cos ...

Question

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

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Solution

The first term of L.H.S. IS $\frac{sin (B - C)}{cos B cos C} = \frac{sin B cos C - cos B sin C}{cos B cos C} = \frac{sin B cos C}{cos B cos C} - \frac{cos B sin C}{cos B cos C} = tan B - tan C$
Similarly, the second term of L.H.S. is $(tan C - tan A)$
and the third term of the L.H.S. is $(tan A - tan B)$
Now L.H.S. $= (tan B - tan C) + (tan C - tan A) + (tan A - tan B) = 0$
Ans: 0

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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}

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\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. 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

Q. If

A + B + C = π

, prove that

(sin A + sin B + sin C) (- sin A + sin B + sin C) (sin A - sin B + sin C) (sin A + sin B - sin C) = 4 {sin}^{2} A {sin}^{2} B {sin}^{2} C

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sin(B−C)cosBcosC+sin(C−A)cosCcosA+sin(A−B)cosAcosB=

The first term of L.H.S. IS sin(B−C)cosBcosC=sinBcosC−cosBsinCcosBcosC=sinBcosCcosBcosC−cosBsinCcosBcosC=tanB−tanCSimilarly, the second term of L.H.S. is (tanC−tanA)and the third term of the L.H.S. is (tanA−tanB)Now L.H.S.=(tanB−tanC)+(tanC−tanA)+(tanA−tanB)=0Ans: 0

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