# Evaluate the following expression without using trigonometric tables:

${\sin ^2}{28^ \circ } + {\sin ^2}{62^ \circ } + {\tan ^2}{38^ \circ } - {\cot ^2}{52^ \circ } + \dfrac{1}{4}{\sec ^2}{30^ \circ }$

Answer

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Hint: Let’s arrange sin terms and tan terms together and use formulae $\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta $ and $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta $ and then simplify further.

Complete step-by-step answer:

Let the value of the given expression be $k$. Then:

$k = {\sin ^2}{28^ \circ } + {\sin ^2}{62^ \circ } + {\tan ^2}{38^ \circ } - {\cot ^2}{52^ \circ } + \dfrac{1}{4}{\sec ^2}{30^ \circ }$

This can be rearranged as:

$k = \left( {{{\sin }^2}{{28}^ \circ } + {{\sin }^2}{{62}^ \circ }} \right) + \left( {{{\tan }^2}{{38}^ \circ } - {{\cot }^2}{{52}^ \circ }} \right) + \dfrac{1}{4}{\sec ^2}{30^ \circ }$

We know that $\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta $ and $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta $, using these two formulae, we’ll get:

$

\Rightarrow k = \left( {{{\sin }^2}{{28}^ \circ } + {{\sin }^2}\left( {{{90}^ \circ } - {{28}^ \circ }} \right)} \right) + \left( {{{\tan }^2}{{38}^ \circ } - {{\cot }^2}\left( {{{90}^ \circ } - {{38}^ \circ }} \right)} \right) + \dfrac{1}{4}{\sec ^2}{30^ \circ }, \\

\Rightarrow k = \left( {{{\sin }^2}{{28}^ \circ } + {{\cos }^2}{{28}^ \circ }} \right) + \left( {{{\tan }^2}{{38}^ \circ } - {{\tan }^2}{{38}^ \circ }} \right) + \dfrac{1}{4}{\sec ^2}{30^ \circ }, \\

\Rightarrow k = \left( {{{\sin }^2}{{28}^ \circ } + {{\cos }^2}{{28}^ \circ }} \right) + 0 + \dfrac{1}{4}{\sec ^2}{30^ \circ } \\

$

As we know that, ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , and $\sec {30^ \circ } = \dfrac{2}{{\sqrt 3 }}$.

Applying these results, we have:

$

\Rightarrow k = 1 + \dfrac{1}{4}{\left( {\dfrac{2}{{\sqrt 3 }}} \right)^2}, \\

\Rightarrow k = 1 + \dfrac{1}{4} \times \dfrac{4}{3}, \\

\Rightarrow k = 1 + \dfrac{1}{3}, \\

\Rightarrow k = \dfrac{4}{3} \\

$

Substituting the value of $k$, we’ll get:

${\sin ^2}{28^ \circ } + {\sin ^2}{62^ \circ } + {\tan ^2}{38^ \circ } - {\cot ^2}{52^ \circ } + \dfrac{1}{4}{\sec ^2}{30^ \circ } = \dfrac{4}{3}$

Thus, the value of the given expression is $\dfrac{4}{3}$.

Note: If any of the trigonometric expression contains $\sin {\theta _1},\cos {\theta _2}$ together or $\tan {\theta _1},\cot {\theta _2}$ together or $\sec {\theta _1},\cos ec{\theta _2}$ together such that ${\theta _1} + {\theta _2} = {90^ \circ }$, then we can always use following results for simplification:

$

\sin \theta = \cos \left( {90 - \theta } \right), \\

\tan \theta = \cot \left( {90 - \theta } \right), \\

\sec \theta = \cos ec\left( {90 - \theta } \right) \\

$

Complete step-by-step answer:

Let the value of the given expression be $k$. Then:

$k = {\sin ^2}{28^ \circ } + {\sin ^2}{62^ \circ } + {\tan ^2}{38^ \circ } - {\cot ^2}{52^ \circ } + \dfrac{1}{4}{\sec ^2}{30^ \circ }$

This can be rearranged as:

$k = \left( {{{\sin }^2}{{28}^ \circ } + {{\sin }^2}{{62}^ \circ }} \right) + \left( {{{\tan }^2}{{38}^ \circ } - {{\cot }^2}{{52}^ \circ }} \right) + \dfrac{1}{4}{\sec ^2}{30^ \circ }$

We know that $\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta $ and $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta $, using these two formulae, we’ll get:

$

\Rightarrow k = \left( {{{\sin }^2}{{28}^ \circ } + {{\sin }^2}\left( {{{90}^ \circ } - {{28}^ \circ }} \right)} \right) + \left( {{{\tan }^2}{{38}^ \circ } - {{\cot }^2}\left( {{{90}^ \circ } - {{38}^ \circ }} \right)} \right) + \dfrac{1}{4}{\sec ^2}{30^ \circ }, \\

\Rightarrow k = \left( {{{\sin }^2}{{28}^ \circ } + {{\cos }^2}{{28}^ \circ }} \right) + \left( {{{\tan }^2}{{38}^ \circ } - {{\tan }^2}{{38}^ \circ }} \right) + \dfrac{1}{4}{\sec ^2}{30^ \circ }, \\

\Rightarrow k = \left( {{{\sin }^2}{{28}^ \circ } + {{\cos }^2}{{28}^ \circ }} \right) + 0 + \dfrac{1}{4}{\sec ^2}{30^ \circ } \\

$

As we know that, ${\sin ^2}\theta + {\cos ^2}\theta = 1$ , and $\sec {30^ \circ } = \dfrac{2}{{\sqrt 3 }}$.

Applying these results, we have:

$

\Rightarrow k = 1 + \dfrac{1}{4}{\left( {\dfrac{2}{{\sqrt 3 }}} \right)^2}, \\

\Rightarrow k = 1 + \dfrac{1}{4} \times \dfrac{4}{3}, \\

\Rightarrow k = 1 + \dfrac{1}{3}, \\

\Rightarrow k = \dfrac{4}{3} \\

$

Substituting the value of $k$, we’ll get:

${\sin ^2}{28^ \circ } + {\sin ^2}{62^ \circ } + {\tan ^2}{38^ \circ } - {\cot ^2}{52^ \circ } + \dfrac{1}{4}{\sec ^2}{30^ \circ } = \dfrac{4}{3}$

Thus, the value of the given expression is $\dfrac{4}{3}$.

Note: If any of the trigonometric expression contains $\sin {\theta _1},\cos {\theta _2}$ together or $\tan {\theta _1},\cot {\theta _2}$ together or $\sec {\theta _1},\cos ec{\theta _2}$ together such that ${\theta _1} + {\theta _2} = {90^ \circ }$, then we can always use following results for simplification:

$

\sin \theta = \cos \left( {90 - \theta } \right), \\

\tan \theta = \cot \left( {90 - \theta } \right), \\

\sec \theta = \cos ec\left( {90 - \theta } \right) \\

$

Last updated date: 25th Sep 2023

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