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 -