Question

Evaluate




cos 58º




sin 32º +




sin 22º




cos 68º –




cos 38º cosec 52º




3 (tan 18º tan 35º tan 60º tan 72º tan 55º)

Answer 1

Answer:

6√3 - 1 / √3

Step-by-step explanation:

Assume question as

Evaluate  

  cos 58 / sin 32 + sin 32 / cos 68 - cos 38 cosec 52 / 3(tan 18 tan 35 tan 60 tan72 tan55)    

         So  Cos (90 - 32)/sin 32 + sin (90 - 68) / cos 68  - cos(90 - 52) cosec 52

       = sin 32 / sin 32 + cos 68 / cos 68 - sin 52 cosec 52 / 3 ( cot 72 tan 72 tan 55 cot 55 tan 60)

    = 1 + 1 - 1 / 3 √3  ( tan 60 = √3)

 =  2 - 1 / 3√3

 = 6√3 - 1 / 3 √3

Answer 2

Answer:

Answer is $\frac{2\sqrt{3}-1}{\sqrt{3}}$

Step-by-step explanation:

Given:

$\frac{cos\,58^{\circ}}{sin\,32^{\circ}}\,+\,\frac{sin\,22^{\circ}}{cos\,68^{\circ}}-\frac{cos\,38^{\circ}\,cosec\,52^{\circ}}{tan\,18^{\circ}\,tan\,25^{\circ}\,tan\,60^{\circ}\,tan\,72^{\circ}\,tan\,65^{\circ}}$

we have to find value of given expression.

As we complimentary angles of trigonometry,

$sin\,x=cos(90^{\circ}-x)\:,\:cos\,x=sin(90^{\circ}-x)\:\:and\:\:tan\,x=cot(90^{\circ}-x)$

Consider,

$\frac{cos\,58^{\circ}}{sin\,32^{\circ}}\,+\,\frac{sin\,22^{\circ}}{cos\,68^{\circ}}-\frac{cos\,38^{\circ}\,cosec\,52^{\circ}}{tan\,18^{\circ}\,tan\,25^{\circ}\,tan\,60^{\circ}\,tan\,72^{\circ}\,tan\,65^{\circ}}$

$\implies\frac{cos\,58^{\circ}}{cos(90^{\circ}-\,32^{\circ})}\,+\,\frac{sin\,22^{\circ}}{sin\,(90^{\circ}-68^{\circ})}-\frac{cos\,38^{\circ}\times\frac{1}{sin\,52^{\circ}}}{tan\,18^{\circ}\,tan\,25^{\circ}\,tan\,60^{\circ}\,cot(90^{\circ}-\,72^{\circ})\,cot(90^{\circ}-\,65^{\circ})}$

$\implies\frac{cos\,58^{\circ}}{cos\,58^{\circ}}\,+\,\frac{sin\,22^{\circ}}{sin\,22^{\circ}}-\frac{cos\,38^{\circ}\times\frac{1}{cos\,(90^{\circ}-52^{\circ})}}{tan\,18^{\circ}\,tan\,25^{\circ}\,tan\,60^{\circ}\,cot\,18^{\circ}\,cot\,25^{\circ}}$

$\implies1\,+\,1-\frac{cos\,38^{\circ}\times\frac{1}{cos\,38^{\circ}}}{tan\,18^{\circ}\,tan\,25^{\circ}\,tan\,60^{\circ}\,\times\frac{1}{tan\,18^{\circ}}\times\frac{1}{\,cot\,25^{\circ}}}$

$\implies1\,+\,1-\frac{1}{tan\,60^{\circ}\,\times1\times1}$

$\implies2-\frac{1}{\sqrt{3}}$     ( ∵ tan 60° = √3 )  

$\implies\frac{2\sqrt{3}-1}{\sqrt{3}}$

Therefore, Answer is $\frac{2\sqrt{3}-1}{\sqrt{3}}$