33 - Trigonometry Questions Answers

We have to prove

tanA + 2tan2A + 4tan4A + 8cot8A = cosecA / secA

or tanA + 2tan2A + 4tan4A + 8cot8A = cotA

or 2tan2A + 4tan4A + 8cot8A = cotA - tanA

or tan2A + 2tan4A + 4cot8A = [1-tan²A]/2tanA

or tan2A + 2tan4A + 4cot8A = cot2A

or 2tan4A + 4cot8A = cot2A - tan2A

or tan4A + 2cot8A = [1-tan²2A]/2tan2A

or tan4A + 2cot8A = cot4A

or 2cot8A = cot4A-tan4A

or cot8A = [1-tan²4A]/2tan4A

it is true so proved

sin(A+B+C+D) = sin(A+B)[cosCcosD-sinCsinD]+cos(A+B)[sinCcosD+cosCsinD]

similarly solve all the terms and add all these terms for getting answer

Submit the correct question, In first bracket / will be there so

((1/(secA.secA-cosA.cosA))-(1/(cosecA.cosecA-sinA.sinA))).sinA.sinA.cosA.cosA

= [cos²A/sin²A(1+cos²A) - sin²A/cos²A(1+sin²A)]sin²Acos²A

= [cos⁴A(1+sin²A) - sin⁴A(1+cos²A)] / [(1+sin²A)(1+cos²A)]

now solve

x²+1 > 2x so cosA>1 which is not possible

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

calculate dx/dθ and dy/dθ

then dy/dx = [2cosθ-2cos2θ]/[-2sinθ+2sin2θ]

now solve

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

and sinC+sinD = 2sin[(C+D)/2]cos[(C-D)/2]

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. 

according to the given condition 

cos^-1x  =  cos^-1(y/2)  +  a

so x = cos ( cos^-1y/2   +  a)

so x = y/2 cosa  -  sin cos^-1y/2 + sina

so x = y/2 cosa  - [1-(y²/4)]^1/2 + sina

so x - sina - y/2 cosa = -[1-(y²/4)]^1/2

now square both side and solve

let sin^-1a = A ,    sin^-1b = B    and    sin^-1c= C 

so  { a√(1-a²) + b√(1-b²) + c√(1-c²)}   = sinAcosA + sinBcosB + sinCcosC   = 1/2 (sin2A+sin2B+sin2C)

                                                                                                          = 1/2 (4sinAsinBsinC)

                                                                                                          = 2abc