Question

sin theta = 1/3 find cos^3 theta • cosec theta + tan theta • sec theta​

Answer 1

$\sf \sin \theta = \frac{1}{3} = \frac{opposite(opp)}{hypotenuse(hyp)}$

$\sf \purple{by \: pythagors \: theorem : }$

$\sf {(hyp)}^{2} = {(opp)}^{2} + {(adjacent)}^{2}$

$\implies \sf {(3)}^{2} = {(1)}^{2} + {(adjacent)}^{2}$

$\implies \sf {(adj)}^{2} = 9 - 1$

$\implies \sf {(adj)}^{2} = 8$

$\therefore \sf adj = 2 \sqrt{2}$

$\implies \sf\cos \theta = \frac{adj}{hyp} = \frac{2 \sqrt{2} }{3}$

$\implies \sf\cosec \theta = \frac{hyp}{opp} = 3$

$\implies \sf \tan \theta = \frac{opp}{adj} = \frac{1}{2 \sqrt{2} }$

$\implies \sf\sec \theta = \frac{hyp}{adj} = \frac{3}{2 \sqrt{2} }$

$\cos ^{3} \theta. \cosec\theta + \tan\theta.\sec\theta$

$\implies [ { (\frac{2 \sqrt{2} }{3} })^{3} \times 3 ]+ [ \frac{2 \sqrt{2} }{3} \times \frac{3}{2 \sqrt{2} } ]$

$\implies [ \frac{16 \sqrt{2} }{9} ]+ [ 1 ]$

$\orange{ \therefore \frac{9 + 16 \sqrt{2} }{9} }$

HOPE IT HELPS!