1) tan9 - tan27- tan63 + tan81 =4
2) sin4 π6 + sin4 3π6 +sin45π6 + sin4 7π6=32
3)find the value of : 4cos20 - √3 cot20
4) find the value of : 2√2 sin10 [sec 52 + cos 40 cos5 - 2sin35]
5)sin212 + sin221 + sin239 + sin248 = 1 + sin29 + sin218
6) if cos(a+b)= 45 and sin(a-b)= 513 ... find the value of 2tan2a
7) find the max nand min value of : 10cos2x - 6sinxcosx + 2sin2x
362) sin4Ï€/6 + 1 + sin4Ï€/6 + sin4Ï€/6 = 1/8 + 1 + 1/8 + 1/8 = 11/8 why 3/2?
366) cos 2A = cos[(A + B) + (A - B)] = cos(A + B).cos(A - B) - sin(A + B).sin(A - B)
Now, cos(A + B) = 4/5 so, sin(A + B) = 3/5
and, sin(A - B) = 5/13 so, cos(A - B) = 12/13
So, cos2A = 4/5.12/13 - 3/5.5/13 = (48 - 15)/65 = 13/65 = 1/5
so, tan2A = 2√6 so, 2tan2A = 4√6 Ans.
1Q.2) take LHS
sin4(Ï€/6)+sin4(3Ï€/6)+sin4(5Ï€/6)+sin4(7Ï€/6)
=sin4(Ï€/2-Ï€/3)+sin4Ï€/2+sin44(Ï€/2+Ï€/3)+sin4(Ï€/2+2Ï€/3)
=cos4Ï€/3+sin4Ï€/2+sin4Ï€/3+sin42Ï€/3
=1+1+9/16≠3/2
even if i have made any calculation mistakes to get 3/2 u need π/8 NOT π/6
