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Mathematics

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Slav-nsk [51]2 years ago

6 0

Give me 8 mins i will answer in the comments

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Prove the identity secxcscx(tanx+cotx)=2+tan^2x+cot^2x

svetlana [45]

Hello,

$sec(x)= \dfrac{1}{cos(x)} \\

cosec(x)= \dfrac{1}{sin(x)} \\

sec(x)*cosec(x)*(tg(x)+cotg(x))=\dfrac{1}{cos(x)}* \dfrac{1}{sin(x)}*( \frac{sin(x)}{cos(x)} +\frac{cos(x)}{sin(x)})\\

= \dfrac{sin^2(x)+cos^2(x)}{sin^2x*cos^2x} \\

= \dfrac{1}{sin^2x*cos^2x} \\

$
==============================================================
$2+tg^2(x)+cotg^2(x)=2+ \dfrac{sin^2x}{cos^2x} + \dfrac{cos^2x}{sin^2x} \\

=2+ \dfrac{sin^4x+cos^4x}{sin^2x*cos^2x} \\

=\dfrac{2*sin^2x*cos^2x+sin^4x+cos^4x}{sin^2x*cos^2x} \\

= \dfrac{(sin^2x+cos^2x)^2}{sin^2x*cos^2x}} \\

= \dfrac{1}{sin^2x*cos^2x}} 
$

8 0

3 years ago

Read 2 more answers

Help and explain pls and thankyouu

defon

Hello,

$\left\{\begin{array}{ccc}2x-y&=&7\\-2x+3y&=&9\\\end{array} \right.\\\\\\\left\{\begin{array}{ccc}2y&=&16\\-2x+3y&=&9\\\end{array} \right.\\\\\\\left\{\begin{array}{ccc}y&=&8\\-2x+3*8&=&9\\\end{array} \right.\\\\\\\left\{\begin{array}{ccc}x&=&7.5\\y&=&8\\\end{array} \right.\\\\$

Answer C

$\left\{\begin{array}{ccc}x-2y-3z&=&5\\x+2y+3z&=&7\\x+z&=&3\\\end{array} \right.\\\\\\\left\{\begin{array}{ccc}2x&=&12\\x+2y+3z&=&7\\x+z&=&3\\\end{array} \right.\\\\\\\left\{\begin{array}{ccc}x&=&6\\2y+3z&=&1\\z&=&-3\\\end{array} \right.\\\\\\\\\left\{\begin{array}{ccc}x&=&6\\y&=&5\\z&=&-3\\\end{array} \right.\\\\\\$