
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
620.7k+ views
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) \\
$
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The draft of the Preamble of the Indian Constitution class 10 social science CBSE

Who gave "Inqilab Zindabad" slogan?

Write a letter to the principal requesting him to grant class 10 english CBSE

Who was Subhash Chandra Bose Why was he called Net class 10 english CBSE

