trignometric-ratios-and-identities-7
Dumb Question:- How should I approach when I look such a question for first time ?
Ans:- Look, here 
was the anle and so we observed that if some how 
could be expressed in tems of 
then our work will be done and so we procceded.
Q. In any triangle ABC prove that
Sin3A cos(B - C) + sin3B cos(C - A) + sin3C cos(A - B) = 3 sinA sinB sinC
Solution:- L.H.S. = sin2A sin(B + C) cos(B - C) + sin2B sin(C + A) cos(C - A) + sin2C sin(A + B) cos(A - B)
                           = 
sin2A (sin2B + sin2C) + sin2B (sin2C + sin2A) + sin2C (sin2A + sin2B)
                          = sin2A (sinB cosB + sinC cosC) + sin2B (sinC cosC + sinA cosA) + sin2C (sinA cosA + sinB cosB)
                          = sinA sinB (sinA cosB + cosA sinB) + sinB sinC (sinB cosC + cosB sinC) + sinC sinA (sinA cosC + cosA sinC)
                          = sinA sinB sin(A + B) + sinB sinC sin(B + C) + sinC sinA sin(A + C)
                          = 3sinA sinB sinC = R.H.S.
Dumb Question:
How did sinA sinB sin(A + B) + sinB sinC sin(B + C) + sinC sinA sin(A + C) became equal to 3 sinA sinB sinC.
Ans:- Since A + B + C = 1800 (Givien that A, B, C are angles of 
)
So, A + B = 1800 -C
and thus sin(A + B) = sin(1800 - C)
                              = sinC
So, sinA sinB sin(A + B) = sinA sinB sinC
Similarily sinB sinC sin(B + C) = sinA sinB sinC
and sinA sinC sin(A + c) = sinA sinB sinC
Q. Show that for all real 
, the expression, a sin2 + b sin cos + c cos2 lies between.
and 
Soluton:- Expression = 
(1 - cos2) + 
sin2 + 
(1 + cos2)
                               = 
+ ( sin2 - 
cos2)
                               
                               
       
                               
But - 1 
sin(2 - 
) 1
maximum value of expression-
                                               
and minimum value of the expression-
                                               
Q-6. If tan + sin = m and tan - sin = n then show that m2 - n2 = 4
Solution:- (m + n) = 2 tan , m - n = 2 sin
m2 - n2 = 4 tan. sin
4 = 4
             = 4 sin
             = 4 sin tan
Dumb Question:-
Why was equated to tan and not - tan ?
Ans:- Yes actually it could have been both, nothin is mentioned here in the question in the question so this ambiquity has arised.
Q-7. Evaluate 
Solution:- Sum = 
                      
                      
Using-
           
                             
                             
                             
                             
                             
So, 
Hard
Q-1. If m2 + m12 + 2mm1 cos = 1
            n2 + n12 + 2nn1 cos = 1 and mn + m1 n1 + (mn1 + m1 n) cos = 1 then prove that m2 + n2 = cosec2
Solution:- m2 + m12 + 2mm1 cos
or (m2cos2 cos2 + m2 sin2) + m12 + 2mm1 cos = 1
or m2 cos2 + 2mm1 cos + m12 = 1 - m2 sin2
or (m cos + m1)2 = 1 - m2 sin2 ………………………………………..(i)
Similaily, n2 + n12 + 2nn1 cos = 1
=> Â Â Â Â Â Â Â Â n2(n cos + n1)2 = 1 - n2 sin2 ………………………………………..(ii)
Finally, mn + m1n1 + (mn1 + m1n cos = 0
=>Â Â Â Â Â Â Â (mn cos2 + mn sin2) + m1n1 cos + m1 n cos = 0
=>Â Â Â Â Â Â Â mn cos2 + m1 n cos + m1 n1 + mn1 cos = -mn sin2
=>Â Â Â Â Â Â Â n cos(m cos + m1) + n1 (m1 + m cos) = - m sin2
=>Â Â Â Â Â Â Â (m cos + m1)(n cos + n1) = - mn sin2
=>Â Â Â Â Â Â Â (m cos + m1).(n cos + n1)2 = m2n2 sin4
=>Â Â Â Â Â Â Â (1 - m2 sin2)(1 - n2 sin2 = m2n2 sin4 Â Â Â Â (Using (i) and (ii))
=>Â Â Â Â Â Â Â 1 - (m2 + n2) sin2 + m2n2 sin4 = m2n2 sin4
=>Â Â Â Â Â Â Â (m2 + n2) sin2 = 1
=>Â Â Â Â Â Â Â m2 + n2 = cosec2