Prove that : from Mathematics Trigonometric Functions

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 Multiple Choice QuestionsShort Answer Type

41.

Prove that: 

2 sin squared open parentheses fraction numerator 11 straight pi over denominator 6 end fraction close parentheses plus cosec squared open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses. cos squared open parentheses straight pi over 3 close parentheses space equals space 3 over 2

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42.

Prove that :

cot squared open parentheses straight pi over 6 close parentheses plus cosec open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses plus 3 tan squared open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses space equals space 6


L.H.S. = cot squared open parentheses straight pi over 6 close parentheses plus cosec open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses plus 3 tan squared open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses

space space space cot squared open parentheses straight pi over 6 close parentheses space equals space cot squared left parenthesis 30 degree right parenthesis space equals space left parenthesis square root of 3 right parenthesis squared space equals space 3

cosec open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses space equals cosec open parentheses fraction numerator 5 cross times 180 degree over denominator 6 end fraction close parentheses space equals space cosec 150 degree space equals space cosec left parenthesis 180 degree minus 30 degree right parenthesis

                   equals space cosec open parentheses straight pi minus straight pi over 6 close parentheses

cosec 150 degree is positive in quadrant II.

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                     tan open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses space equals space tan 210 degree space equals space tan left parenthesis 180 degree plus 30 degree right parenthesis space equals space tan open parentheses straight pi plus straight pi over 6 close parentheses

tan 210 degree is +ve in quadrant III.

∴       tan open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses space equals space tan open parentheses straight pi over 6 close parentheses space equals space fraction numerator 1 over denominator square root of 3 end fraction

         L.H.S. = cot squared open parentheses straight pi over 6 close parentheses plus cosec open parentheses fraction numerator 5 straight pi over denominator 6 end fraction close parentheses plus 3 tan squared open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses
                  equals 3 plus 2 plus 3 cross times open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses squared space equals space 3 plus 2 plus 1 space equals 6 space equals space straight R. straight H. straight S.
                   


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43.

Prove that:

2 sin squared open parentheses fraction numerator 3 straight pi over denominator 4 end fraction close parentheses plus 2 cos squared open parentheses straight pi over 4 close parentheses plus 2 sec squared open parentheses fraction numerator 7 straight pi over denominator 6 end fraction close parentheses space equals 10

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 Multiple Choice QuestionsLong Answer Type

44.

Prove that:

fraction numerator sin left parenthesis 180 degree plus straight theta right parenthesis. space cos left parenthesis 90 degree plus straight theta right parenthesis space tan left parenthesis 270 degree minus straight theta right parenthesis space cot left parenthesis 360 degree minus straight theta right parenthesis over denominator sin left parenthesis 360 degree minus straight theta right parenthesis space cos left parenthesis 360 degree plus straight theta right parenthesis space cosec left parenthesis 0 degree minus straight theta right parenthesis space sin left parenthesis 270 degree plus straight theta right parenthesis end fraction equals 1

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 Multiple Choice QuestionsShort Answer Type

45.

Show that:  <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>

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46.

Show that:  cos 24 degree plus cos 55 degree plus cos 125 degree plus cos 204 degree plus cos 300 degree space equals space 1 half

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47.

Show that:  sin squared straight pi over 8 plus sin squared fraction numerator 3 straight pi over denominator 8 end fraction plus sin squared fraction numerator 5 straight pi over denominator 8 end fraction plus sin squared fraction numerator 7 straight pi over denominator 8 end fraction space equals space 2

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48.

Solve for x, when sec open parentheses straight pi over 2 plus straight A close parentheses plus xsinA space tan open parentheses straight pi over 2 plus straight A close parentheses space equals space cos open parentheses straight pi over 2 plus straight A close parentheses. Also, find the value of x, when <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>

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49. In ∆ABC, prove that:

space space space cos space left parenthesis straight A plus straight B right parenthesis plus cos space straight C space equals space 0
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50. In ∆ABC, prove that:

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#6 {main}</pre>


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