# 10 - Trigonometrical Equations Questions Answers

CosecA+CotA=5/2 than TanA=?
Asked By: NITESH KUMAR 4 year ago
If A+B+C=180, then secA(cosBcosC-sinBsinC) is equal to
Asked By: SATYAM KUMAR SINGH 4 year ago

Prove - tanA+cotA can never be equal to 3/2

Asked By: SHIVAM GUPTA 5 year ago
If(1+4x^2)cosA=4x,show that cosecA+cotA=1+2x÷1-2x
Asked By: SANKET GUPTA 6 year ago
2cos7x/cos(3)+sin(3)>2^cos2x
Asked By: AARSHI 6 year ago

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

it is cos3 or cos3x

(tanx)^2(tan3x)^2(tan4x)=(tanx)^2-(tan3x) ^2 + tan4x.
Asked By: AARSHI 6 year ago

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

## (B) 5/3

Asked By: ARJUN 7 year ago

sinA + sinB + sinC = 4cosA/2.cosB/2.cosC/2

Asked By: AVI GARG 7 year ago

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.

A regular hexagon and a regular dodecagon are inscribed in the same circle. if the side of the dodecagon is (√3-1), then the side of hexagon is

Asked By: HEMA RANI 7 year ago

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

If     a sinx + b cos(x+θ) + b cos(x-θ) = d, then the minimum value of |cos θ| is equal to :

sir i have expanded cos(x+θ) & cos(x-θ) and after that i got the equation: (d sec x - a tanx)/2b = cos θ. now , can i write  minimum value of (d sec x -a tanx) = (d²-a²)^(1/2).. as all the options contain (d²-a²)^(1/2).......or do tell the other hint to solve it..

Asked By: AMIT DAS 8 year ago

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)