It is
From
canceling the
This can be expanded through the tangent angle addition formula:
#tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)#
Thus,
#tan(x+y)=(tanx+tany)/(1-tanxtany)#
The tangent addition formula can be found using the sine and cosine angle addition formulas.
#sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta#
#cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta#
Since
#tan(alpha+beta)=sin(alpha+beta)/cos(alpha+beta)=(sinalphacosbeta+cosalphasinbeta)/(cosalphacosbeta-sinalphasinbeta)#
This can be written in terms of tangent by dividing both the numerator and denominator by
#tan(alpha+beta)=((sinalphacosbeta+cosalphasinbeta)/(cosalphacosbeta))/((cosalphacosbeta-sinalphasinbeta)/(cosalphacosbeta))=(sinalpha/cosalpha(cosbeta/cosbeta)+sinbeta/cosbeta(cosalpha/cosalpha))/(cosalpha/cosalpha(cosbeta/cosbeta)-sinalpha/cosalpha(sinbeta/cosbeta))#
Final round of simplification yields:
#tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)#
Euler formula :
We have to write in terms of
We get
First of all we have to convert these two numbers into trigonometric forms.
If
Magnitude of a complex number
Let
Magnitude of
Angle of
Let
Magnitude of
Angle of
Now,
Here we have every thing present but if here directly substitute the values the word would be messy for find
We know that:
This is your final answer.
You can also do it by another method.
By firstly dividing the complex numbers and then changing it to trigonometric form, which is much easier than this.
First of all let's simplify the given number
Multiply and divide by the conjugate of the complex number present in the denominator i.e
Let
Magnitude of
Angle of
First of all we have to convert these two numbers into trigonometric forms.
If
Magnitude of a complex number
Let
Magnitude of
Angle of
Let
Magnitude of
Angle of
Now,
Here we have every thing present but if here directly substitute the values the word would be messy for find
We know that:
This is your final answer.
You can also do it by another method.
By firstly dividing the complex numbers and then changing it to trigonometric form, which is much easier than this.
First of all let's simplify the given number
Multiply and divide by the conjugate of the complex number present in the denominator i.e
Let
Magnitude of
Angle of
Area
The area of a triangle
In this case
So
Area
Write in terms of
#=costheta-cos^2theta+cos^2theta/sin^2theta#
Find a common denominator.
#=(costhetasin^2theta)/sin^2theta-(cos^2thetasin^2theta)/sin^2theta+cos^2theta/sin^2theta#
Combine.
#=(costhetasin^2theta-cos^2thetasin^2theta+cos^2theta)/sin^2theta#
The following simplification may seem unecessary, but is actually relevant. Its purpose will become clear in the following step.
#=(sintheta(color(blue)(costhetasintheta))-color(green)(cos^2theta)sin^2theta+color(green)(cos^2theta))/sin^2theta#
Use the following identities:
#=(sintheta((sin2theta)/2)-(1-sin^2theta)sin^2theta+(1-sin^2theta))/sin^2theta#
#=((sinthetasin2theta)/2-sin^2theta+sin^4theta+1-sin^2theta)/sin^2theta#
#=((sinthetasin2theta)/2-2sin^2theta+sin^4theta+1)/sin^2theta#
#=(2sin^4theta-4sin^2theta+sinthetasin2theta+2)/(2sin^2theta)#
As
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Determine
Using Sine Rule and
Determine h using
Determine area
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Sum internal angles of a triangle is
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using
But
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
but
The
The
Remembering our trigonometry, the vertical component of a vector is given by
and the horizontal component by
in the polar coordinate
Hence:
The
The
In this case,