Question

Demostrar las identidades trigonométricas

Answer 1

1.)

${cosA ctgA+senA=}\\{cosA( \frac{cosA}{senA} )+senA=}\\{ \frac{cos^2A}{senA} +senA=}\\{ \frac{cos^2A+sen^2A}{senA}=}\\{ \frac{1}{senA}=}\\{cscA}$

2.)
${cscA secA-tanA=}\\{ \frac{1}{senA} ( \frac{1}{cosA} )- \frac{senA}{cosA} =}\\{ \frac{1}{senAcosA} -\frac{senA}{cosA}=}\\{ \frac{1-sen^2A}{senAcosA}=}\\{ \frac{cos^2A}{senAcosA}=}\\{\frac{cosA}{senA}=}\\{ctgA}$

4.)
${ senAcosAtanA=}\\{( senA)( cosA} )(\frac{senA}{cosA}) =}\\{ sen^2A=}\\{1-cos^2A}$

5.)
${ ctgAsec^2A-ctgA=}\\{( \frac{cosA}{senA})( \frac{1}{cos^2A} )-\frac{cosA}{senA} =}\\{ \frac{1}{senAcosA}-\frac{cosA}{senA}=}\\{\frac{1-cos^2A}{senAcosA}=}\\{\frac{sen^2A}{senAcosA}=}\\{\frac{senA}{cosA}=}\\{tanA}$

3.)
${secAtanAcscA=}\\{( \frac{1}{cosA})(\frac{senA}{cosA})(\frac{1}{senA})=}\\{\frac{1}{cos^2A}=}\\{\frac{sen^2A+cos^2A}{cos^2A}=}\\{\frac{sen^2A}{cos^2A}+\frac{cos^2A}{cos^2A}}\\{tan^2A+1}$
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Salu2. :)