1. for all θ∈R find the value of tan²θ sec²θ(cot²θ-cos²θ) 2. if x=y cos2π/3=z cos4π/3 than find the value of xy+yz+zx 3. if cosecθ - sinθ =m and secθ + cosecθ =n , eliminate θ

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

1. `tan^2 theta sec^2 theta (cot^2 theta - cos^2 theta)`

`tan^2 theta sec^2 theta cot^2 theta - tan^2 theta sec^2 theta cos^2 theta`

Sine tan and cot are reciprocals, and sec and cos are reciprocal you get:

`=sec^2 theta - tan^2 theta` and by the Pythagorean identities

=1

2. Given `x=y cos((2pi)/3)=z cos((4pi)/3)`

Substituting the values for `cos((2pi)/3)=1/2, cos((4pi)/3)=-1/2` we get:

`x=1/2 y = -1/2 z`

Then y=2x; z=-2x so

`xy+yz+zx=x(2x)+(2x)(-2x)+x(-2x)=2x^2-4x^2-2x^2=-4x^2`

Sources:
gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

Part 3 of the question is answered with the stated condition of the question i.e.  secθ + cosecθ =n.

Please see attachment.

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kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

For the First two questions answers given by "embizee" are the simple one .....no need other solutions .They are good.

Now the

3)

Given 

m = cosec `theta` - sin `theta` = `(1 - sin^2theta)/(sin theta)`

                                 = `(cos^2theta)/(sin theta)` and

n = `sec theta- cos theta`  = `(1 - cos^2theta)/cos theta`

                                    = `(sin^2theta)/cos theta.`

So

`m/n = (cos^3 theta)/(sin^3 theta)` = `cot^3 theta`  

and

`n/m = (sin^3 theta )/(cos^3 theta) = tan^3 theta` .

Thus

`tan^2 theta= (m/n)^(2/3)` and `(cot^2 theta)= (n/m)^(2/3)` .

Next, we square both original equations and use the formula

`cosec theta` `sin theta`  = `sec theta` `cos theta`  = 1, we get

`m^2` = `cosec^2theta+ sin ^2theta - 2`

     = `(cosec^2theta - 1)+ sin ^2theta - 1`

     = `cot ^2theta+ sin ^2theta- 1 `

and

`n^2` = `sec ^2 theta` `+` `cos ^2 theta`  `- 2` = `tan ^2 theta`  `+ cos ^2 theta`  `- 1` .

we know `cosec ^2 theta- 1` = `cot ^2 theta`

and `sec ^2 theta- 1` = `tan ^2 theta.`

If we move the 1s to the other sides and add, we get

`m^2 + n^2 + 2 ` = `cot ^2 theta+ tan ^2 theta+ sin ^2 theta+ cos ^2 theta`

 = `cot ^2 theta+ tan ^2 theta+ 1` .

when we Subtract 1 from both sides, and we get

`m^2 + n^2 + 1` `= ` `(m/n)^(2/3) + (n/m)^(2/3)` .  

:)

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gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

Posted on

In the question it is given that secθ + cosecθ =n , while it has been solved assuming n=secθ -Cosθ

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