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

2 answers:

8 0

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}} 
$

KiRa [710]3 years ago

3 0

sec(x)csc(x)[tan(x) + cot(x)] = 2 + tan²(x) + cot²(x)
sec(x)csc(x)[tan(x)] + sec(x)csc(x)[cot(x)] = 2 + tan²(x) + cot²(x)
sec²(x) + csc²(x) = 2 + tan²(x) + cot²(x)
sec²(x) + csc²(x) = 1 + 1 + tan²(x) + cot²(x)
sec²(x) + csc²(x) = 1 + tan²(x) + 1 + cot²(x)
sec²(x) + csc²(x) = sec²(x) + csc²(x)

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8. Is the following statement true or false? Justify your conclusion. If a and b are nonnegative real numbers and a + b =0, then

atroni [7]

Answer:

If $a + b = 0$ then $a = 0$ and $b =0$

a | b | a + b (answer)

0 | 0 | 0

0 | 1 | 1

0 | 2 | 2

1 | 0 | 1

2 | 0 | 2

1 | 1 | 2

2 | 1 | 3

Step-by-step explanation:

Considering the following conditions for the real numbers:

$a\geq 0\\b\geq 0$

Following the rules of these in-equations, it is possible to deduce:

$a + b \geq 0$

Then, if the proposed statement is:

$a + b = 0$

The conditions above shall comply the requirements established, but first, analyzing the statement:

If $a \geq 0$ and $b \geq 0$ then $a + b \geq a$ , $a + b \geq b$ and $a + b \geq 0$ .

If $a = 0$ and b a non negative real number, then $a + b \geq b$ , but because to $a = 0$ , then $a + b = b$ . Due to the commutative property of sums, the same behavior will be presented if $b = 0$ and a a non negative real number.