Question

4(sin^4 60 + cos^4 30) -3(tan^2 60-tan^2 45)+5cos^2 45

Answer 1

Solution:

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Given & To Find:

The value of,

$4(sin^4 60 + cos^4 30) -3(tan^2 60-tan^2 45)+5cos^2 45$

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As we know that,

$sin 60 = \frac{ \sqrt{3} }{2}$,

$cos 30 = \frac{ \sqrt{3} }{2}$,

$tan 60 = \sqrt{3}$,

tan 45 = 1,

$cos 45 = \frac{1}{ \sqrt{2} }$,.

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So,

$4(sin^4 60 + cos^4 30) -3(tan^2 60-tan^2 45)+5cos^2 45$

=> $4(( \frac{ \sqrt{3} }{2} )^4 + ( \frac{ \sqrt{3} }{2}) ^4) -3( (\sqrt{3})^2 - (1)^2 )+5( \frac{1}{ \sqrt{2} } )$

=> $4(( \frac{9}{4} + \frac{9}{4} ) -3( (3 - 1 )+( \frac{5}{ \sqrt{2} } )$

=> $4(( \frac{18}{4} ) -3( 2 )+( \frac{5}{ \sqrt{2} } )$

=> $18 -6 +( \frac{5}{ \sqrt{2} } )$

=> $12 - \frac{5}{ \sqrt{2} }$

=> $\frac{ 12\sqrt{2}-5 }{ \sqrt{2} }$

=> $\frac{24 - 5 \sqrt{2} }{2}$

=> $12 - 2.5 \sqrt{2}$

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Answer 2

4 (sin^4 60°+cos^4 30°)-3 ( tan^2 60°- tan^2 45°)+5 (cos^2 45°)

Solution:-

=> 4 {(root of 3/2)^4+(root of 3/2)^4}-3 {( root of 3)^2-1}+5 (1/ root of 2)^×

=> 4(9/16+9/16)-3 (3-1)+5 (1/2)

=> 4 (18/16)-3 (2)+5/2

=> 18/4-6+5/2

=> 18-24+10/4

=> -6+10/4

=> 4/4

=> 1