# 10 - Trigonometrical Equations Questions Answers

**Asked By: NITESH KUMAR**3 year ago

**read solutions ( 1 ) | submit your answer**

**Asked By: SATYAM KUMAR SINGH**3 year ago

**read solutions ( 1 ) | submit your answer**

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

**Asked By: SHIVAM GUPTA**4 year ago

**read solutions ( 1 ) | submit your answer**

**Asked By: SANKET GUPTA**5 year ago

**submit your answer**

**Asked By: AARSHI**5 year ago

**submit your answer**

**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

**Asked By: AARSHI**5 year ago

**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

tan^{2}x tan^{2}3x tan4x = tan^{2}x - tan^{2}3x + tan4x

so tan4x = [tan^{2}x - tan^{2}3x] / [tan^{2}x tan^{2}3x-1]

now split RHS terms by formula x^{2}-y^{2}= (x+y)(x-y)

so RHS = tan4xtan2x

now solve

##
If cos^{2}θ -sin^{2}θ=1/3, value of (cos^{4}θ-sin^{4}θ+1) is:

## (A) 4/3

## (B) 5/3

**Asked By: ARJUN**6 year ago

**read solutions ( 12 ) | submit your answer**

__sinA + sinB + sinC __= 4cosA/2.cosB/2.cosC/2

**Asked By: AVI GARG**6 year ago

**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

**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

**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 sec^{2}x = dy/dx

on comparing with 0 we get d tanx = a secx or sinx = a/d so secx = d/√(d^{2}-a^{2}) and tanx = a/√(d^{2}-a^{2})

so minimum of given expression = [d^{2}/ √(d^{2}-a^{2})] - [a^{2}/√(d^{2}-a^{2})] = √(d^{2}-a^{2})