Question

5 cos2 60 + 4 sec2 30 - tan2 45 / sin2 30 + cos 30

Answer 1

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Answer 2

(5 cos2 60 + 4 sec2 30 - tan2 45 ) / (sin2 30 + cos 30) =  67/(3+6√3)

Explanation:

TRIGONOMETRIC RATIOS :

                   0°        30°           45°          60°         90°  

Sin$\theta$             0          1/2         1/√2        √3/2          1

cos$\theta$              1         √3/2       1/√2          1/2           0

tan$\theta$              0         1/√3         1            √3            undefined

cot$\theta$           not         √3            1            1/√3           0

            defined

Sec$\theta$             1           2/√3       √2             2              not defined

cosec$\theta$        not          2            √2          2/√3           1  

               defined

Given:

(5 cos2 60 + 4 sec2 30 - tan2 45 ) / (sin2 30 + cos 30)

From that, Tabulation find the ratios value

cos 60° =     1/2

sec 30° =     2/√3  

tan 45°  =      1

sin 30°  =      1/2

cos 30° =     √3/2

==>(5 cos2 60 + 4 sec2 30 - tan2 45 )

cos²60° = (1/2)²

cos²60° = 1/4

sec²30°   =( 2/√3 )²

sec²30°   = 4/3

tan²45°  =  (1)²

tan²45°  =  1

==>(sin2 30 + cos 30)

sin²30° = ( 1/2)²

sin²30° = 1/4

cos 30° =  √3/2

Substitute all the values in the given question,

==> (5 cos2 60 + 4 sec2 30 - tan2 45 ) / (sin2 30 + cos 30)

==> (5(1/4) + 4 (4/3) - 1) / (1/4 +√3/2 )

==> (5/4 + 16/3 -1) / (1/4 +√3/2 )

The denominator is different

so, Taking LCM

==> (5(3)/4(3) + 16(4)/3(4) - 1(12)/12) / (1/4 + √3(2)/2(2) )

==>  (15/12 +  64/12 - 12/12) / (1/4 +2√3/4)

==> (3/12 +64/12) / ((1+2√3)/4)

==> (67/12 ) / (1+2√3)/4

==> (67/12 ) × 4/(1+2√3)

==> (67/3) × 1/(1+2√3)

==> 67/3(1+2√3)

==> 67/(3+6√3)

==> (5 cos2 60 + 4 sec2 30 - tan2 45 ) / (sin2 30 + cos 30) =  67/(3+6√3)