trignometric-ratios-and-identities-8

Q-2. Show that
Solution: Let  (Note )

or
or 2 sin² cos2 = 3 sin - 4 sin³
or 4 cos.cos2 = 3 - 4 sin² ( sin 0)
or 4 cos.(2 cos² - 1)
This is a third degree equation in cos and each of satisfied it.
roots of (1) are
= sum of roots = - ................................................. (2)
       ............................................. (3)
       .................................................................................... (4)
Now
                                                  



      [Using (2), (3), (4)]

Hence

Dumb Question:- How was equated to .
Ans:-
                                                       =


Q-3. If A + B + C = , sinA sinB sinC = P and cosA cosB cosC = 2 then otain the cubic equation whoose roots are tanA, tanB and tanC.
Solution:- The required cubic equation is
    (x - tanA)(x - tanB)(x - tanC) = 0
or x³ - (tanA + tanB + tanC) x² + x(tanA tanB + tanB tanC + tanC tanA) - tanA tanB tanC = 0 ...............................................(1)
Now tanA. tanB. tanC = .........................................................................(2)
Also tan(A + B) = tan( - C) = - tanC
or = - tanC
or tanA + tanB + tanC = tanA tanB tanC = ..............................................................(3)    [Using (2)]
Now sin²A + sin²B + sin²C
= [3 - (cos2A + cos2B + cos2C)]
= [3 - {2 cos(A + B).cos(A - B) + 2 cos²C - 1}]
= [4 - {- 2cosC cos(A - B) + 2cos²C}]
= [4 + 2 cosC{cos(A - B) + cos(A + B)}]
= 2(1 + cosA cosB cosC) = 2(1 + q)
sin²A sin²B sin²C (cosec²B cosec²C + cosec²C cosec²A + cosec²A cosec²B ) = 2(1 + q)
or p²{(1 + cot²B)(1 + cot²C) + (1 + cot²C) (1 + cot²A) + (1 + cot²A) (1 + cot²B)} = 2(1 + q)
or p²{3 + 2 (cot²A + cot²B + cot²C) + cot²B cot²C + cot²C cot²A + cot²A cot²B)} = 2(1 + q)
or p²[3 + 2{cotA + cotB + cotC)² - 2(cotA cotB + cotB cotC + cotC cotA)} + {(cotB cotC + cotC cotA + cotA cotB)² - 2 cotA cotB x cotC(cotA + CotB + cotC)}] = 2(1 + q)
or p² [3 + 2{(cotA + cotB + cotC)² - 2
+ {1 - 2.(cotA + cotB + cotC)}] = 2 (1 + q) ..........................................(4)
[cotA. cotB = 1 if A + B + C = ]
Let y = tanA tanB + tanB tanC + tanC tanA
         = tanA tanB tanC(cotC + cotA + cotB)
         = (cotA + cotB + cotC )
cotA + cotB + cotC = ........................................................(5)
(4) gives
or
or -2(1 + q)
or q²y² - q²y - (1 - q) = 0
y =
........................................................... (6)
From (1), (2), (3), (4), (5) and (6),
x³ - x² + x - = 0   where y =
or x³ x² - x - = 0    when y = -
The required equation is
    qx³ - px² + (q + 1)x - p = 0
or qx³ - px² - x - p = 0


Dumb Question:-
How does suddenly we started evaluating sin²A + sin²B + sin²C ?
Ans:- The aim that we were trying to achievc at that point of time was to calculat the value of tanA tanB + tanB tanC + tan A tanC.
Now when we try to evaluate this we realize that cotA + cotB + cotC needs to evaluated and thus we tried to make a equation is terms of cotA + cotB +cotC
So, all this factors were take in account and then only the sin²A + sin²B + sin²C was eqaluated.

Key words:
(1) Trignometry
(2) Quadrant
(3) Sin
(4) Cos
(5) Tan
(6) Cot
(7) Sec
(8) Cosec
(9) Allied angles.
(10) Compound angles.
(11) Prelendicular
(12) Hypotenuse
(13) Base