Question

If tan theta is =8/15, evaluate: sin theta - 1 / cos theta (1 - cos theta)

Answer 1

Solution:-

Given

$\rm \to \: \tan \theta = \dfrac{8}{15}$

To find

$\to \rm \dfrac{ \sin \theta - 1}{ \cos\theta (1 - \cos\theta)}$

We know that

$\rm \to \: \tan \theta = \dfrac{8}{15} = \dfrac{p}{b}$

Using pythagoras theorem

$\rm \to \: {h}^{2} = {p}^{2} + {b}^{2}$

$\rm \: {h}^{2} = 8 {}^{2} + 15 {}^{2}$

$\rm {h}^{2} = 64 + 225$

$\rm {h}^{2} = 289$

$\rm \: h {}^{} = \sqrt{289}$

$\rm \: h = 17$

we know that

$\rm \sin \theta = \dfrac{p}{h} = \dfrac{8}{17} \: \: \: and \: \: \cos \theta = \dfrac{b}{h} = \dfrac{15}{17}$

Now put the value on

$\to \rm \dfrac{ \sin \theta - 1}{ \cos\theta (1 - \cos\theta)}$

$\to \rm \: \dfrac{ \dfrac{8}{17} - 1}{ \dfrac{15}{17} \bigg(1 - \dfrac{15}{17} \bigg) }$

$\to \rm \dfrac{ \dfrac{8 - 17}{17} }{ \dfrac{15}{17} \bigg( \dfrac{17 - 15}{17} \bigg) }$

$\rm \to \: \dfrac{ \dfrac{ - 9}{17} }{ \dfrac{15}{17} \times \dfrac{2}{17} }$

$\rm \to \: \dfrac{ \dfrac{ - 9}{17} }{ \dfrac{30}{289} }$

$\to \rm \dfrac{ - 9}{17} \times \dfrac{289}{ 30}$

$\rm \to \: \dfrac{ - 9 \times 17}{30}$

$\rm \to \dfrac{ - 153}{30}$