1(a)
given tan2theta=-20/21
Let tantheta =x
So using the identity tan2theta=(2tantheta)/(1-tan^2theta) we get
-20/21=(2x)/(1-x^2)
=>-10/21=x/(1-x^2)
=>10x^2-21x-10=0
=>10x^2-25x+4x-10=0
=>5x(2x-5)+2(2x-5)=0
=>(2x-5)(5x+2)=0
So x=5/2 and x=-2/5
Hence tantheta=5/2 and tantheta=-2/5
`
1(b)
given tan2theta=-36/77
Let tantheta =x
So using the identity tan2theta=(2tantheta)/(1-tan^2theta) we get
36/77=(2x)/(1-x^2)
=>18/77=x/(1-x^2)
=>18x^2+77x-18=0
=>18x^2+81x-4x-18=0
=>9x(2x+9)-2(2x+9)=0
=>(2x+9)(9x-2)=0
So x=-9/2 and x=2/9
Hence tantheta=-9/2 and tantheta=2/9
`
2(a)
cos2theta+5costheta=2
=>2cos^2theta-1+5costheta-2=0
=>2cos^2theta+5costheta-3=0
=>2cos^2theta+6costheta-costheta-3=0
=>(2costheta-1)(costheta+3)=0
So costheta=-3->"not possible"
when
(2costheta-1)=0
=>costheta=1/2=cos60^@=cos(360-60)^@=cos300^@
Hence theta=60^@ and theta=300^@
`
2(b)
2sec2x-cot2x=tan2x
=>2sec2x=tan2x+cot2x
=>2sec2x=(sin2x)/(cos2x)+(cos2x)/(sin2x)
=>2sec2x=(sin^2 2x+cos^2 2x)/(sin2xcos2x)=csc2xsec2x
=>2sec2x-csc2xsec2x=0
=>sec2x(2-csc2x)=0
sec2x=0-> "not possible"
when (2-csc2x)=0
sin2x=1/2=sin30^@=sin(180-30)=sin150^@
Hence x= 15^@ and x= 75^@
2(c)
2tan(theta/2)+3tantheta=0
=>2tan(theta/2)+(6tan(theta/2))/(1-tan^2(theta/2))=0
=>2tan(theta/2)(1+3/(1-tan^2(theta/2)))=0
=>2tan(theta/2)(4-tan^2(theta/2))=0
when
tan(theta/2)=0=tan(o^@)=tan180^@
Hence theta = 0^@ and theta = 360^@
(4-tan^2(theta/2))=0
=>tan(theta/2)=+-2
when
=>tan(theta/2)=+2=tan63.43^@
=>theta =126.86^@
when
=>tan(theta/2)=-2=-tan63.43^@=tan(180-63.43)
=>tan(theta/2)=tan116.57^@
=>theta =233.14^@
