How can you prove: secθ = cscθtanθ

1 answer:

3 0

$\sec \theta=\csc \theta\tan \theta\\\\ \dfrac{1}{\cos \theta}=\dfrac{1}{\sin \theta}\cdot\dfrac{\sin \theta}{\cos \theta}\\\\ \dfrac{1}{\cos \theta}=\dfrac{1}{\cos \theta}$

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gtnhenbr [62]

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$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$