Show prove the below identity? $\frac{1}{cos 290} + \frac{1}{\sqrt{3} sin 250} = \frac{4}{\sqrt{3}}$ $1/cos290 + 1/(sqrt3sin250) = 4/sqrt3$

Question

Show prove the below identity? $\frac{1}{cos 290} + \frac{1}{\sqrt{3} sin 250} = \frac{4}{\sqrt{3}}$ $1/cos290 + 1/(sqrt3sin250) = 4/sqrt3$

$\frac{1}{cos 290} + \frac{1}{\sqrt{3} sin 250} \equiv \frac{4}{\sqrt{3}}$ $1/cos290 + 1/(sqrt3sin250) -= 4/sqrt3$

2 Answers

Impact of this question

4439 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-11T03:51:39

$L H S = \frac{1}{{cos 290}^{\circ}} + \frac{1}{\sqrt{3} {sin 250}^{\circ}}$ $LHS=1/(cos290^@)+1/(sqrt3sin250^@)$

$= \frac{1}{{cos (360 - 70)}^{\circ}} + \frac{1}{\sqrt{3} {sin (180 + 70)}^{\circ}}$ $=1/(cos(360-70)^@)+1/(sqrt3sin(180+70)^@)$

$= \frac{1}{{cos 70}^{\circ}} - \frac{1}{\sqrt{3} {sin 70}^{\circ}}$ $=1/(cos70^@)-1/(sqrt3sin70^@)$

$= \frac{\sqrt{3} {sin 70}^{\circ} - {cos 70}^{\circ}}{\sqrt{3} {sin 70}^{\circ} {cos 70}^{\circ}}$ $=(sqrt3sin70^@-cos70^@)/(sqrt3sin70^@cos70^@)$

$= \frac{1}{\sqrt{3}} ⎡ ⎢ ⎣ \frac{2 {\sqrt{3} {sin 70}^{\circ} - {cos 70}^{\circ}}}{2 {sin 70}^{\circ} {cos 70}^{\circ}} ⎤ ⎥ ⎦$ $=1/sqrt3[(2{sqrt3sin70^@-cos70^@})/(2sin70^@cos70^@)]$

$= \frac{1}{\sqrt{3}} ⎡ ⎢ ⎢ ⎣ \frac{2 \cdot 2 {{sin 70}^{\circ} \cdot (\frac{\sqrt{3}}{2}) - {cos 70}^{\circ} \cdot (\frac{1}{2})}}{{sin 140}^{\circ}} ⎤ ⎥ ⎥ ⎦$ $=1/sqrt3[(2*2{sin70^@*(sqrt3/2)-cos70^@*(1/2)})/(sin140^@)]$

$= \frac{1}{\sqrt{3}} [\frac{4 {{sin 70}^{\circ} \cdot {cos 30}^{\circ} - {cos 70}^{\circ} \cdot {sin 30}^{\circ}}}{{sin (180 - 40)}^{\circ}}]$ $=1/sqrt3[(4{sin70^@*cos30^@-cos70^@*sin30^@})/(sin(180-40)^@)]$

$=1/sqrt3[(4{sin(70-30)^@})/(sin40^@)]=1/sqrt3[(4{cancel(sin40^@)})/cancel((sin40^@))]=4/sqrt3=RHS$

NOTE that $cos(360-A)^@=cosA and sin(180+A)^@=-sinA$

Answer 2 · 2018-05-11T03:58:49

$1/cos290 + 1/(sqrt3sin250)$

$=1/cos(270+20) + 1/(sqrt3sin(270-20))$

$=1/sin20 - 1/(sqrt3cos20)$

$=[(sqrt3cos20-sin20)/(sqrt3sin20cos20)]$

$=2/sqrt3[(sqrt3/2cos20-1/2sin20)/(sin20cos20)]$

$=4/sqrt3[(sin60cos20-cos60sin20)/(2sin20cos20)]$

$=4/sqrt3[sin(60-20)/(2sin20cos20)]$

$=4/sqrt3[sin40/sin40]$

$= 4/sqrt3$

Science

Math

Humanities

... and beyond

Show prove the below identity? $\frac{1}{cos 290} + \frac{1}{\sqrt{3} sin 250} = \frac{4}{\sqrt{3}}$ $1/cos290 + 1/(sqrt3sin250) = 4/sqrt3$

$\frac{1}{cos 290} + \frac{1}{\sqrt{3} sin 250} \equiv \frac{4}{\sqrt{3}}$ $1/cos290 + 1/(sqrt3sin250) -= 4/sqrt3$

2 Answers

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Show prove the below identity? 1/cos290 + 1/(sqrt3sin250) = 4/sqrt3 1cos290+1√3sin250=4√31/cos290 + 1/(sqrt3sin250) = 4/sqrt3

1/cos290 + 1/(sqrt3sin250) -= 4/sqrt31cos290+1√3sin250≡4√31/cos290 + 1/(sqrt3sin250) -= 4/sqrt3

2 Answers

Related questions

Impact of this question

Show prove the below identity? $\frac{1}{cos 290} + \frac{1}{\sqrt{3} sin 250} = \frac{4}{\sqrt{3}}$ $1/cos290 + 1/(sqrt3sin250) = 4/sqrt3$

$\frac{1}{cos 290} + \frac{1}{\sqrt{3} sin 250} \equiv \frac{4}{\sqrt{3}}$ $1/cos290 + 1/(sqrt3sin250) -= 4/sqrt3$