trignometric-ratios-and-identities-5

Conditional Identities.
If A + B + C = 180^o then .
(i) sin 2A + sin 2B + sin 2c = 4sinA sin B sinC.

Why ?

A + B + C = 180^o
So, A + B = 180^o - C
sin(A + B) = sin C
cos (A + B) = - cos C
sin 2A + sin 2B + sin 2C = 2 sin (A + B) cos (A - B) + 2sin C cos C.
= 2sin C cos (A - B) + 2sin (- cos (A + B))
= 2sin C (cos (A - B) - cos (A + B))
= 2sin C (2sin A sin B)
= 4sin A sin B sin C .

(ii) cos 2A + cos 2B + cos 2C = - 1 - 4cos A cosB cosC

(iii) sinA + sinB + sinC = 4 cos

(iv) cosA + cosB + cosC = 1 + 1 + 4sin

(v) tanA + tanB + tanC =tanA tanB tanC

Why ?
tan (A + B + C) =
=
=
So, since tan (A + B + C) = tan (180⁰) = 0
So, tanA + tanB + tanC - tanA + tanB + tanC = 0
=> tanA + tanB + tanC = tanA + tanB + tanC.

(vi) cotA + cotB + cotA + cotC + cotB cotC = 1

(vii)

(viii)

Illustration 10:
If A + B + C = 180 ^o then prove that
sin² A + sin² C = 2 + 2 cosA cosB cosC.
Ans:- sin²A + sin²B + sin²C = 1/2 (1 - cos²A + 1 - cos²B + 1 - cos²C)
= (3 - (cos2A + cos2B + cos2C))
= [3 - (- 1 - 4cosA cosB cosC)]
= [4 + 4cosA cosB cosC]
= cosA cosB cosC
= 2 + 2 cosA cosB cosC

1. Question - If sin + cosec = 2 then what is the value of sin² + coses².
   Solution:- sin² + cosesc² = (sin + cosec)² - 2sin cosec
   = (1 + 1)² - 2.1 = 2
   
   
2. Prove that 2(sin⁶ + cos⁶) - 3(sin⁴ + cos⁴ ) + 1 = 0
   Solution:- L.H.S. 2 [ (sin² + cos²)³ - 3sin²cos²(sin²+ cos²) ] -[ (sin² + cos²)³ - 3sin²cos²] + 1
   = 2 [ 1 - 3 sin²cos²] - 3 [ 1 - 2sin²cos²] + 1
   = 0
   
   
3. Prove that for all real
   
   Solution sin² + cos⁴ - cos² + 1
   
   =
   
   =
   
   Also, sin² + cos² = cos⁴- cos² + 1
   
   = cos²(cos² - 1) + 1
   = 1 - sin² cos² 1
   
   [ becaise sin² cos² 0]
   
   
4. Evaluate sin 78 ^o - sin66^o - sin42 ^o + sin6^o
   = (sin 78^o - sin42^o) - ( sin66^o - sin6^o)
   = 2cos 60^o sin18^o - 2cos36^o. sin30^o
   
   = sin18^o - cos36^o
   =
   = -
   
   (5) If a then find the maximum value of sinA sinB .
   Solution:- sinA sinB = 2 sinA sinB
   = [ cos (A - B) - cos (A + B)]
   = [cos (A - b) - cos 90^o]
   = cos (A - B)
   So maximum valuie of sinA sinB =
   
   (6) Prove that cosec 20^o - sec 20^o = 4
   
   Soluction- L.H.S. =
   
   =
   
   =4.
   
   = 4.
   = 4.
   
   = 4 = R.H.S.
   
   (7) Find the value of,
   
   When | tanA | < | and | A | is acure,
   
   Solution:-
   =>
   
   =>
   
   => and in this interval casA > sinA }
   
   => => cotA
   
   When |tanA| < 1 and |A| is acute.
   
   Q. Find a and b such that for all x, a
   Soluction:- 3 cosx + 5 sin(x - ) = 3 cosx + 5 sinx cos - 5 cosx sin
                                                       = (3 - 5/2)cosx + 5 sinx
                                                       = cosx + sinx
         a =
   and b =
   
   (9) Find the maximum and minimum value of cos² - 6 sin cos + 3 sin² + 2
   Soluction:- cos² - 6 sin.cos + 3 sin² + 2
   = (1 - sin²) - 3 sin2 + 3 sin² + 2
   = 2 sin² - 3 sin2 + 3
   = (1 - cos2) - 3sin2 + 3
   = 4 - (cos2 + 3 sin2) …………………………………………. (i)
   as we have -
   
   or 4 -