10 - Trigonometrical Equations Questions Answers

CosecA+CotA=5/2 than TanA=?
Asked By: NITESH KUMAR 3 year ago
is this question helpfull: 21 4 read solutions ( 1 ) | submit your answer
If A+B+C=180, then secA(cosBcosC-sinBsinC) is equal to
Asked By: SATYAM KUMAR SINGH 3 year ago
is this question helpfull: 5 3 read solutions ( 1 ) | submit your answer

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

Asked By: SHIVAM GUPTA 4 year ago
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If(1+4x^2)cosA=4x,show that cosecA+cotA=1+2x÷1-2x
Asked By: SANKET GUPTA 4 year ago
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2cos7x/cos(3)+sin(3)>2^cos2x
Asked By: AARSHI 5 year ago
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Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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 5 year ago
is this question helpfull: 12 1 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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

If cos2θ -sin2θ=1/3, value of (cos4θ-sin4θ+1) is:

(A) 4/3

(B) 5/3

Asked By: ARJUN 6 year ago
is this question helpfull: 6 2 read solutions ( 12 ) | submit your answer

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

Asked By: AVI GARG 6 year ago
is this question helpfull: 3 0 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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 6 year ago
is this question helpfull: 2 0 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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 6 year ago
is this question helpfull: 2 0 submit your answer
Answer Strategies and trick by Manish sir (it will help you to solve it by yourself)

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)   

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