Conditional Trigonometric Identities


Here, you can find the solutions of exercise 1 of conditional trigonometric identities. The questions are relating to 'tangent' and 'cotangent'. Solutions have been provided in PDF (Portable Document Format) for the viewers ease. You can download the solutions if you wish to.

All Exercises Of This Chapter:
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5

Exercise 4

If A, B and C are the angles of a triangle, prove that:

$1. tan2A+tan2B+tan2C = tan2A.tan2B.tan2C$

$2. tan\frac{A}{2}.tan\frac{B}{2} + tan\frac{B}{2}.tan\frac{C}{2} + tan\frac{A}{2}.tan\frac{C}{2} = 1$

$3. cotA.cotB + cotB.cotC + cotC.cotA = 1$

$4. cot\frac{A}{2}.cot\frac{B}{2}.cot\frac{C}{2} = cot\frac{A}{2} + cot\frac{B}{2}+cot\frac{C}{2} $

$5. tan(A+B-C) +tan(B+C-A) +tan(C+A-B) = tan(A+B-C).tan(B+C-A).tan(C+A-B)$



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4. If A, B, C are the angles of a triangle, prove that:

a) sin (B + C - A) + sin (C + A - B) + sin (A + B -C) = 4 sinA sinB sinC

Solution:

$$\rm A + B + C = \pi$$

$$\rm (A + B) = \pi - C$$

$$\rm \cos(A + B) = cos (\pi - C) = - \cos C$$

$$\rm sin (A +B) = sin ( \pi - C) = sin C$$

LHS

$\rm \sin ( B + C - A) + \sin (C + A - B) + \sin (A + B - C)$

$\rm = \sin ( A + B + C - 2A) + \sin (C + A + B - 2B) + \sin (A + B + C - 2C)$

$\rm = \sin (\pi - 2A) + \sin (\pi - 2B) + \sin (\pi - 2C)$

$\rm = \sin 2A + \sin 2B + \sin 2C$

$\rm = 2 \sin (A+B) \cos (A-B) + 2 \sin C \cos C$

$\rm = 2 \sin C \cos (A - B) + 2 \sin C \cos C$

$\rm = 2 \sin C \{ \cos ( A - B) + \cos C \}$

$\rm = 2 \sin C \{ \cos (A - B) - \cos (A + B) \}$

$\rm = 2 \sin C \times 2 \sin A \sin B$

$\rm = 4 \sin A \sin B \sin C$

= RHS


b) cos ( B + C - A) + cos(C + A -B) + cos (A + B - C) = 1 + 4 cosA cosB cos C

Solution:

$$\rm A + B + C = \pi$$

$$\rm (A + B) = \pi - C$$

$$\rm \cos(A + B) = cos (\pi - C) = - \cos C$$

$$\rm sin (A +B) = sin ( \pi - C) = sin C$$

LHS

$\rm \cos ( B + C - A) + \cos (C + A -B) + \cos (A + B - C)$

$\rm = \cos (B + C + A - 2A) + \cos ( C + A + B - 2B) + \cos (A + B + C - 2C)$

$\rm = \cos (\pi - 2A) + \cos (\pi - 2B) + \cos (\pi - 2C)$

$\rm = - \cos 2A - \cos 2B - \cos 2C$

$\rm = - ( \cos 2A + \cos 2B + \cos 2C )$

$\rm = - ( 2 \cos (A + B) \cos ( A - B) + \cos 2 C )$

$\rm = - ( - 2 \cos C \cos (A - B) + 2 \cos^2 C - 1)$

$\rm = 1 - 2 \cos C ( \cos C - \cos (A - B))$

$\rm = 1 - 2 \cos C \{ - \cos ( A+ B) - \cos ( A - B) \} $

$\rm = 1 - 2 \cos C ( - 2 \cos A \cos B)$

$\rm = 1 + 4 \cos A \cos B \cos C$

= RHS


c) sinA cosB cosC + sinB cosC cosA + sinC cosA cosB = sinA sinB sinC

Solution:

$$\rm A + B + C = \pi$$

$$\rm (A + B) = \pi - C$$

$$\rm \cos(A + B) = cos (\pi - C) = - \cos C$$

$$\rm sin (A +B) = sin ( \pi - C) = sin C$$

LHS

$\rm \sin A  \cos B \cos C + \sin B \cos C \cos A + \sin C \cos A \cos B$

$\rm = \cos C ( \sin A \cos B + \sin B \cos A) + \sin C \cos A \cos B$

$\rm = \cos C \sin (A +B) + \sin C \cos A \cos B$

$\rm = \cos C \sin C + \sin C \cos A \cos B$

$\rm = \sin C ( \cos C + \cos A \cos B)$

$\rm = \sin C ( - \cos (A + B) + \cos A \cos B)$

$\rm = \sin C \times \sin A \sin B$

$\rm = \sin A \sin B \sin C$

= RHS


d) cosA sinB sinC + cosB sinC sinA + cosC sinA sinB = 1 + cosA cosB cosC

Solution:

$$\rm A + B + C = \pi$$

$$\rm (A + B) = \pi - C$$

$$\rm \cos(A + B) = cos (\pi - C) = - \cos C$$

$$\rm sin (A +B) = sin ( \pi - C) = sin C$$

LHS

$\rm \cos A \sin B \sin C + \cos B \sin C \sin A + \cos C \sin A \sin B$

$\rm = \sin C ( \cos A \sin B + \cos B \sin A) + \cos C \sin A \sin B$

$\rm = \sin C \sin ( A + B) + \cos C \sin A \sin B$

$\rm = \sin C \sin C + \cos C \sin A \sin B$

$\rm = 1 - \cos^2 C + \cos C \sin A \sin B$

$\rm = 1 - \cos C ( \cos C - \sin A \sin B)$

$\rm = 1 - \cos C ( - \cos (A + B) - \sin A \sin B) $

$\rm = 1 + \cos C ( \cos (A + B) + \sin A \sin B ) $

$\rm = 1 + \cos C \times \cos A \cos B$

$\rm = 1 + \cos A \cos B \cos C$

= RHS