Math, asked by meow1231, 8 days ago

prove that:
1+sec0-tan0/tan0/1+sec0+tan0 = 1-sin0/cos0

Answers

Answered by BrainlyPARCHO
1

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

LHS=sin0-cos0+1/sin0+cos0-1

on dividing by cos0 up and down,then

tan0+se0-1/tan0-sec0+1

= sec0+tan0-(sec20-tan20)/tan0-sec0+1

= sec0+tan0-{(sec0+tan0)(sec0-tan0)}/tan0-sec0+1

=sec0+tan0(1-sec0+tan0)/tan0-sec0+1

=sec0+tan0*(sec0-tan0)/(sec0-tan0)

=(sec20-tan20)/(sec0-tan0)

=1/sec0-tan0)

=RHS

Answered by BrainlyPARCHO
0

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

LHS=sin0-cos0+1/sin0+cos0-1

on dividing by cos0 up and down,then

tan0+se0-1/tan0-sec0+1

= sec0+tan0-(sec20-tan20)/tan0-sec0+1

= sec0+tan0-{(sec0+tan0)(sec0-tan0)}/tan0-sec0+1

=sec0+tan0(1-sec0+tan0)/tan0-sec0+1

=sec0+tan0*(sec0-tan0)/(sec0-tan0)

=(sec20-tan20)/(sec0-tan0)

=1/sec0-tan0)

=RHS

Answered by BrainlyPARCHO
0

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

LHS=sin0-cos0+1/sin0+cos0-1

on dividing by cos0 up and down,then

tan0+se0-1/tan0-sec0+1

= sec0+tan0-(sec20-tan20)/tan0-sec0+1

= sec0+tan0-{(sec0+tan0)(sec0-tan0)}/tan0-sec0+1

=sec0+tan0(1-sec0+tan0)/tan0-sec0+1

=sec0+tan0*(sec0-tan0)/(sec0-tan0)

=(sec20-tan20)/(sec0-tan0)

=1/sec0-tan0)

=RHS

Answered by BrainlyPARCHO
0

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

LHS=sin0-cos0+1/sin0+cos0-1

on dividing by cos0 up and down,then

tan0+se0-1/tan0-sec0+1

= sec0+tan0-(sec20-tan20)/tan0-sec0+1

= sec0+tan0-{(sec0+tan0)(sec0-tan0)}/tan0-sec0+1

=sec0+tan0(1-sec0+tan0)/tan0-sec0+1

=sec0+tan0*(sec0-tan0)/(sec0-tan0)

=(sec20-tan20)/(sec0-tan0)

=1/sec0-tan0)

=RHS

Answered by BrainlyPARCHO
0

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

LHS=sin0-cos0+1/sin0+cos0-1

on dividing by cos0 up and down,then

tan0+se0-1/tan0-sec0+1

= sec0+tan0-(sec20-tan20)/tan0-sec0+1

= sec0+tan0-{(sec0+tan0)(sec0-tan0)}/tan0-sec0+1

=sec0+tan0(1-sec0+tan0)/tan0-sec0+1

=sec0+tan0*(sec0-tan0)/(sec0-tan0)

=(sec20-tan20)/(sec0-tan0)

=1/sec0-tan0)

=RHS

Answered by BrainlyPARCHO
0

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

LHS=sin0-cos0+1/sin0+cos0-1

on dividing by cos0 up and down,then

tan0+se0-1/tan0-sec0+1

= sec0+tan0-(sec20-tan20)/tan0-sec0+1

= sec0+tan0-{(sec0+tan0)(sec0-tan0)}/tan0-sec0+1

=sec0+tan0(1-sec0+tan0)/tan0-sec0+1

=sec0+tan0*(sec0-tan0)/(sec0-tan0)

=(sec20-tan20)/(sec0-tan0)

=1/sec0-tan0)

=RHS

Answered by BrainlyPARCHO
0

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

LHS=sin0-cos0+1/sin0+cos0-1

on dividing by cos0 up and down,then

tan0+se0-1/tan0-sec0+1

= sec0+tan0-(sec20-tan20)/tan0-sec0+1

= sec0+tan0-{(sec0+tan0)(sec0-tan0)}/tan0-sec0+1

=sec0+tan0(1-sec0+tan0)/tan0-sec0+1

=sec0+tan0*(sec0-tan0)/(sec0-tan0)

=(sec20-tan20)/(sec0-tan0)

=1/sec0-tan0)

=RHS

Answered by BrainlyPARCHO
1

\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}

LHS=sin0-cos0+1/sin0+cos0-1

on dividing by cos0 up and down,then

tan0+se0-1/tan0-sec0+1

= sec0+tan0-(sec20-tan20)/tan0-sec0+1

= sec0+tan0-{(sec0+tan0)(sec0-tan0)}/tan0-sec0+1

=sec0+tan0(1-sec0+tan0)/tan0-sec0+1

=sec0+tan0*(sec0-tan0)/(sec0-tan0)

=(sec20-tan20)/(sec0-tan0)

=1/sec0-tan0)

=RHS

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