Question

AYUDA CON ESTOS PROCEDIMIENTOS POR FAVOR

Answer 1

trigonometria:

8)

sabiendo que $sec^{2}A=tan^{2}A+1$

$\frac{tan^{2}A}{sec^{2}A}+ \frac{1}{sec^{2}A} =1\\\\\frac{tan^{2}A+1}{sec^{2}A} =1\\\\\frac{sec^{2}A}{sec^{2}A} =1\\1=1$

9)

sabiendo que $tan^{2}\alpha =\frac{sen^{2}\alpha }{cos^{2}\alpha }$

sabiendo que  $sen^{2}\alpha +cos^{2}\alpha =1$

$tan^{2}\alpha -sen^{2}\alpha =sen^{2}\alpha *tan^{2}\alpha \\\frac{sen^{2}\alpha }{cos^{2}\alpha } - sen^{2}\alpha = sen^{2}\alpha * \frac{sen^{2}\alpha }{cos^{2}\alpha }$

$\frac{sen^{2}\alpha -sen^{2}\alpha*cos^{2}\alpha }{cos^{2}\alpha } =sen^{2}\alpha*\frac{sen^{2}\alpha}{cos^{2}\alpha}$

$\frac{sen^{2}\alpha(1 -cos^{2}\alpha) }{cos^{2}\alpha } =\frac{sen^{2}\alpha*sen^{2}\alpha}{cos^{2}\alpha}$

$\frac{sen^{2}\alpha*sen^{2}\alpha }{cos^{2}\alpha } =\frac{sen^{2}\alpha*sen^{2}\alpha}{cos^{2}\alpha}$

10)

sabiendo que   $csc\alpha =\frac{1}{sen\alpha }$

sabiendo que $cot\alpha =\frac{1}{tan\alpha }$

sabiendo que  $sen^{2}\alpha +cos^{2}\alpha =1$

$csc\alpha - sen\alpha =cos\alpha *cot\alpha$

$\frac{1}{sen\alpha } -sen\alpha =cos\alpha *\frac{1}{tan\alpha }$

$\frac{1}{sen\alpha } -\frac{sen^{2}\alpha }{sen\alpha } =cos\alpha *\frac{1}{\frac{sen\alpha }{cos\alpha } }$

$\frac{1-sen^{2}\alpha}{sen\alpha} =cos\alpha *\frac{cos\alpha }{sen\alpha }$

$\frac{1-sen^{2}\alpha}{sen\alpha} =\frac{cos^{2}\alpha }{sen\alpha }$

$\frac{cos^{2}\alpha}{sen\alpha} =\frac{cos^{2}\alpha }{sen\alpha }$