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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