Question

if tan theta =8/15, find the value of sin theta + cos theta/cos theta(1-cos theta)?

Answer 1

i hope it is help for you

Answer 2

$Given \: tan \:\theta = \frac{8}{15} \: --(1)$

$sec^{2} \theta \\= 1 + tan^{2} \theta \\= 1 + \Big( \frac{8}{15}\Big)^{2} \\= 1 + \frac{64}{225} \\= \frac{225 + 64 }{225 } \\= \frac{289}{225} \\= \frac{17^{2}}{15^{2}}$

$Now, sec \theta = \sqrt{\Big(\frac{17^{2}}{15^{2}}\Big)} \\= \frac{17}{15}$

$Cos \theta = \frac{15}{17}\: --(2)$

$Value \:of\:\frac{sin \theta + cos \theta }{cos \theta( 1 - cos \theta )}$

$Dividing \:numerator \:and \:denominator \:by \\cos \theta , \: we \:get$

$= \frac{\frac{sin \theta + cos \theta}{cos\theta} }{\frac{cos \theta( 1 - cos \theta )}{cos \theta }}$

$= \frac{ \frac{sin \theta}{cos \theta} + \frac{cos \theta}{cos \theta }}{ 1 - cos \theta }\\= \frac{tan \theta - 1 }{ 1 - cos \theta} \\= \frac{ \frac{8}{15}+ 1}{ 1 - \frac{15}{17}}\\= \frac{\frac{8 + 15}{15}}{\frac{17-15}{17}}\\= \frac{\frac{23}{15}}{\frac{2}{17}} \\= \frac{23}{15} \times \frac{17}{2}\\= \frac{371}{30}$

Therefore.,

$\red {Value \:of\:\frac{sin \theta + cos \theta }{cos \theta( 1 - cos \theta )}}\green {= \frac{371}{30} }$

•••♪