6 - Trigonometrical Equations Questions Answers

Please submit what we have to do and also correct the question 

it is cos3 or cos3x 

I dont know what the question is but if the question is to solve this eq. then the method will be as given below

tan²x tan²3x tan4x = tan²x  - tan²3x + tan4x

so tan4x =  [tan²x  - tan²3x] / [tan²x tan²3x-1]

now split RHS terms by formula x²-y²= (x+y)(x-y)

so RHS = tan4xtan2x 

now solve

sinA + sinB + sinC = 2sin[(A+B)/2]cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2sin[(π-C)/2]cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2cosC/2cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2cosC/2* [cos[(A-B)/2]+[cos(A+B)/2]                     here sinC/2 = cos(A+B)/2

now use cosC + cosD = 2 cos[(C+D)/2]cos[(C-D)/2]

You should remember that the given formula is based on a triangle. 

for dodecagon sin(2π/40) = (a/2)/r     so    r = (a/2)/ sin(π/20)

now for hexagon sin(2π/12) = (x/2)/r       so x/2 = r sin(π/6)  =  (a/2) sin(π/6)/ sin(π/20)  

so  x  = (√3-1) (1/2) / ((√5-1)/4)                      here sin(π/20) = (√5-1)/4

or x = 4(√3-1) / 2(√5-1) 

now solve it for further simplification

Dear amit your starting steps are correct 

now let us consider that   d secx - a tanx  = y

on differentiating we will get d secx tanx - a sec²x = dy/dx

on comparing with 0 we get d tanx = a secx       or               sinx = a/d               so         secx = d/√(d²-a²)           and   tanx = a/√(d²-a²)

so minimum of given expression = [d²/ √(d²-a²)] - [a²/√(d²-a²)]  =        √(d²-a²)