Question 1170774
How to prove

1.

{{{sin(theta) = cos(theta)*tan(theta)}}}

start with right side

{{{cos(theta)*tan(theta)}}}...........use identity {{{tan(theta)=sin(theta)/cos(theta)}}}

={{{cos(theta)(sin(theta)/cos(theta))}}}.........simplify

={{{cross(cos(theta) )(sin(theta)/cross(cos(theta)))}}}

= {{{sin(theta)}}}



2. 

{{{1/cos(theta)=tan(theta)/sin(theta)}}}

start with right side

{{{tan(theta)/sin(theta)}}}.........use identity {{{tan(theta)=sin(theta)/cos(theta)}}}


={{{(sin(theta)/cos(theta))/sin(theta)}}}


={{{sin(theta)/(cos(theta)*sin(theta))}}}


={{{cross(sin(theta))1/(cos(theta)*cross(sin(theta)))}}}


={{{1/cos(theta)}}}



3. 

{{{cos(2theta) tan(2theta)= sin(2theta)}}}

start with left side

{{{cos(2theta) tan(2theta)}}}.........use identities {{{cos(2theta)=cos^2(theta) - sin^2(theta)}}} and {{{tan(2theta)= sin(2theta)/cos(2theta)}}}

={{{(cos^2(theta) - sin^2(theta))(sin(2theta)/cos(2theta))}}}....use  again {{{cos(2theta)=cos^2(theta) - sin^2(theta)}}}

={{{(cos^2(theta) - sin^2(theta))(sin(2theta)/(cos^2(theta) - sin^2(theta)))}}}

={{{(cross(cos^2(theta) - sin^2(theta)))(sin(2theta)/(cross(cos^2(theta) - sin^2(theta))))}}}

={{{sin(2theta)}}}


4. 

{{{sin(theta)tan(theta) + cos(theta) =1/cos(theta)}}}

start with left side

{{{sin(theta)tan(theta) + cos(theta) }}}.....use identity {{{tan(theta)=sin(theta)/cos(theta)}}}

={{{sin(theta)(sin(theta)/cos(theta)) + cos(theta) }}}

={{{sin^2(theta)/cos(theta) + cos(theta)}}} ....use identity {{{sin^2(theta)=1-cos^2(theta)}}}

={{{(1-cos^2(theta))/cos(theta) + cos(theta) }}}

={{{(1-cos^2(theta))/cos(theta) + cos^2(theta)/cos(theta) }}}

={{{(1-cos^2(theta) + cos^2(theta))/cos(theta) }}}

={{{(1-cross(cos^2(theta)) +cross( cos^2(theta)))/cos(theta) }}}

={{{1/cos(theta) }}}


5. 

{{{sin(theta) (1 + tan(theta) ) = tan(theta) (sin(theta)  + cos(theta) )}}}

start with left side

{{{sin(theta) (1 + tan(theta) )}}} ....use identity {{{tan(theta)=sin(theta)/cos(theta)}}}

={{{sin(theta) (1 + sin(theta)/cos(theta) )}}}

={{{sin(theta) (cos(theta) /cos(theta) + sin(theta)/cos(theta) )}}}

={{{sin(theta) ((cos(theta) + sin(theta))/cos(theta) )}}}....factor out {{{1/cos(theta) }}}

={{{sin(theta)(1/cos(theta) ) (cos(theta) + sin(theta))}}}

={{{(sin(theta)/cos(theta) ) (cos(theta) + sin(theta))}}}

={{{tan(theta)  (cos(theta) + sin(theta))}}}