How do you show that $csctheta = (1+ sec theta)/(tan theta + sintheta)$ ?

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Answer 1 · 2017-02-19T02:09:39

I like to convert everything to sine and cosine and go from there. Use the following identities to do so:

• $csctheta = 1/sintheta$
• $sectheta = 1/costheta$
• $tantheta = sintheta/costheta$

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$1/sintheta = (1+ 1/costheta)/(sintheta/costheta + sin theta)$

$1/sintheta = ((costheta + 1)/costheta)/((sin theta + sinthetacostheta)/costheta)$

$1/sintheta = (costheta + 1)/costheta * costheta/(sin theta + sinthetacostheta)$

$1/sintheta = (costheta+ 1)/costheta * costheta/(sintheta(1 + costheta))$

$1/sintheta = 1/sintheta$

$LHS = RHS$

This identity has been proved.

Hopefully this helps!

Answer 2 · 2017-02-19T02:09:42

$RHS= (1+sec theta )/ (tanx + sinx)$

$= (1+sec theta )/ (sinx/cosx + sinx)$

$= (1+sec theta )/ (sinxsecx + sinx)$

$= (1+sec theta )/ (sinx(secx + 1))$

$=1/sinx=cscx=LHS$

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How do you show that csctheta = (1+ sec theta)/(tan theta + sintheta)csctheta = (1+ sec theta)/(tan theta + sintheta)?