Question

Evaluate:
A. \[\dfrac{{\sin {{30}^ \circ } + \tan {{45}^ \circ } - \cos ec{{60}^ \circ }}}{{\sec {{30}^ \circ } + \cos {{60}^ \circ } + \cot {{45}^ \circ }}}\]
B. \[\dfrac{{5{{\cos }^2}{{60}^ \circ } + 4{{\sec }^2}{{30}^ \circ } - {{\tan }^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{30}^ \circ }}}\]

Answer 1

Hint: These are some basic questions of trigonometry we will be using only one trigonometric formula \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and we will also be needing values of \[\sin {30^ \circ } = \dfrac{1}{2},\tan {45^ \circ } = 1,\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}\] to solve the following questions.

Complete step by step answer:
We must know some properties of trigonometry before attempting the question, those are as follows:
\[\begin{array}{l}
\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)\\
\tan \theta = \cot \left( {{{90}^ \circ } - \theta } \right)\\
\sec \theta = \cos ec\left( {{{90}^ \circ } - \theta } \right)
\end{array}\]
Also that
\[\begin{array}{l}
\cos ec\theta = \dfrac{1}{{\sin \theta }}\\
\sec \theta = \dfrac{1}{{\cos \theta }}\\
\cot \theta = \dfrac{1}{{\tan \theta }}
\end{array}\]

A. Clearly \[\cos ec{60^ \circ } = \dfrac{1}{{\sin {{60}^ \circ }}}\& \cos ec{60^ \circ } = \sec {30^ \circ }\]
Also \[\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}\] therefore \[\cos ec{60^ \circ } = \sec {30^ \circ } = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{2}{{\sqrt 3 }}\]
Putting all of these in \[\dfrac{{\sin {{30}^ \circ } + \tan {{45}^ \circ } - \cos ec{{60}^ \circ }}}{{\sec {{30}^ \circ } + \cos {{60}^ \circ } + \cot {{45}^ \circ }}}\]
We will get
\[\begin{array}{l}
 = \dfrac{{\dfrac{1}{2} + 1 - \dfrac{2}{{\sqrt 3 }}}}{{\dfrac{2}{{\sqrt 3 }} + \dfrac{1}{2} + 1}}\\
 = \dfrac{{\dfrac{{\sqrt 3 + 2\sqrt 3 - 4}}{{2\sqrt 3 }}}}{{\dfrac{{4 + \sqrt 3 + 2\sqrt 3 }}{{2\sqrt 3 }}}}\\
 = \dfrac{{\sqrt 3 + 2\sqrt 3 - 4}}{{4 + \sqrt 3 + 2\sqrt 3 }}\\
 = \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}
\end{array}\]
By rationalising this we will get
\[\begin{array}{l}
\therefore \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}}\\
 = \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}} \times \dfrac{{3\sqrt 3 - 4}}{{3\sqrt 3 - 4}}\\
 = \dfrac{{{{\left( {3\sqrt 3 - 4} \right)}^2}}}{{{{\left( {3\sqrt 3 } \right)}^2} - {{\left( 4 \right)}^2}}}\\
 = \dfrac{{27 - 24\sqrt 3 + 16}}{{27 - 16}}\\
 = \dfrac{{43 - 24\sqrt 3 }}{{11}}
\end{array}\]

B. Here we are given the question \[\dfrac{{5{{\cos }^2}{{60}^ \circ } + 4{{\sec }^2}{{30}^ \circ } - {{\tan }^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {{\cos }^2}{{30}^ \circ }}}\]
By solving we can get
\[\begin{array}{l}
 = \dfrac{{5 \times {{\left( {\dfrac{1}{2}} \right)}^2} + 4 \times {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2} - {1^2}}}{1}\\
 = \dfrac{5}{4} + 4 \times \dfrac{3}{4} - 1\\
 = \dfrac{5}{4} + 3 - 1\\
 = \dfrac{5}{4} + 2\\
 = \dfrac{{13}}{4}
\end{array}\]

Note: It must be noted that \[\dfrac{1}{{\sin A}} = \cos ecA\] also many students usually put \[{{{\sin }^2}A + {{\cos }^2}A}\] in place of 1 which makes the calculation much longer.Just like \[{{{\sin }^2}A + {{\cos }^2}A = 1}\] we also have \[{\sec ^2}A - {\tan ^2}A = 1\& \cos e{c^2}A - {\cot ^2}A = 1\] Also note that \[\sin \theta = \dfrac{p}{h},\cos \theta = \dfrac{b}{h},\tan \theta = \dfrac{p}{b},\cos ec\theta = \dfrac{h}{p},\sec \theta = \dfrac{h}{b},\cot \theta = \dfrac{b}{p}\] where p, b, h represents perpendicular, base and height of a right angled triangle respectively.