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Find the value of these trigonometric quantities:
\[
  {\text{(i) }}\sin {60^0}\cos {30^0} + \sin {360^0}\cos {60^0} \\
  {\text{(ii) 2}}{\left( {{\text{tan}}{{45}^0}} \right)^2} + {\left( {\cos {{30}^0}} \right)^2} - {\left( {\sin {{60}^0}} \right)^2} \\
  {\text{(iii) }}\dfrac{{\cos {{45}^0}}}{{\sec {{30}^0} + {\text{cosec}}{{30}^0}}} \\
  {\text{(iv) }}\dfrac{{\sin {{30}^0} + \tan {{45}^0} - {\text{cosec}}{{60}^0}}}{{\sec {{30}^0} + \cos {{60}^0} + \cot {{45}^0}}} \\
  {\text{(v) }}\dfrac{{5{{\left( {\cos {{60}^0}} \right)}^2} + 4{{\left( {\sec {{30}^0}} \right)}^2} - {{\left( {\tan {{45}^0}} \right)}^2}}}{{{{\left( {\sin {{30}^0}} \right)}^2} + {{\left( {\cos {{30}^0}} \right)}^2}}} \\
 \]

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Last updated date: 12th Jul 2024
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Answer
VerifiedVerified
448.8k+ views
Hint: Here, we will be using a trigonometric table for the known values of the trigonometric functions corresponding to some angles.

Complete step-by-step answer:
Since, we know all the values for these predefined trigonometric functions used. Hence, by substituting we will get
\[
  {\text{(i) }}\sin {60^0}\cos {30^0} + \sin {360^0}\cos {60^0} = \dfrac{{\sqrt 3 }}{2} \times \dfrac{{\sqrt 3 }}{2} + 0 \times \dfrac{1}{2} = \dfrac{3}{4} + 0 = \dfrac{3}{4}. \\
  {\text{(ii) 2}}{\left( {{\text{tan}}{{45}^0}} \right)^2} + {\left( {\cos {{30}^0}} \right)^2} - {\left( {\sin {{60}^0}} \right)^2} = 2 \times {1^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} = 2. \\
  {\text{(iii) }}\dfrac{{\cos {{45}^0}}}{{\sec {{30}^0} + {\text{cosec}}{{30}^0}}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{2}{{\sqrt 3 }} + 2}} = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{{2 + 2\sqrt 3 }}{{\sqrt 3 }}}} = \dfrac{{\sqrt 3 }}{{2\sqrt 2 \left( {1 + \sqrt 3 } \right)}}. \\
  {\text{(iv) }}\dfrac{{\sin {{30}^0} + \tan {{45}^0} - {\text{cosec}}{{60}^0}}}{{\sec {{30}^0} + \cos {{60}^0} + \cot {{45}^0}}} = \dfrac{{\dfrac{1}{2} + 1 - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{2}{{\sqrt 3 }} + \dfrac{1}{2} + 1}} = \dfrac{{\dfrac{{\sqrt 3 + 2\sqrt 3 - 4}}{{2\sqrt 3 }}}}{{\dfrac{{4 + \sqrt 3 + 2\sqrt 3 }}{{2\sqrt 3 }}}} = \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}. \\
  {\text{(v) }}\dfrac{{5{{\left( {\cos {{60}^0}} \right)}^2} + 4{{\left( {\sec {{30}^0}} \right)}^2} - {{\left( {\tan {{45}^0}} \right)}^2}}}{{{{\left( {\sin {{30}^0}} \right)}^2} + {{\left( {\cos {{30}^0}} \right)}^2}}} = \dfrac{{5 \times {{\left( {\dfrac{1}{2}} \right)}^2} + 4 \times {{\left( {\dfrac{2}{{\sqrt 3 }}} \right)}^2} - {1^2}}}{{{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}} = \dfrac{{5 \times \left( {\dfrac{1}{4}} \right) + 4 \times \left( {\dfrac{4}{3}} \right) - 1}}{{\dfrac{1}{4} + \dfrac{3}{4}}} \\
   = \dfrac{{\dfrac{5}{4} + \dfrac{{16}}{3} - 1}}{{\dfrac{4}{4}}} = \dfrac{{\dfrac{{15 + 64 - 12}}{{12}}}}{1} = \dfrac{{67}}{{12}}. \\
 \]
Note: These types of problems are easily solved by substituting the values of the trigonometric functions at some predefined angles which are taken from the trigonometric table.