Evaluate, without using trigonometric tables:
\[\dfrac{{{{\sin }^2}\theta + {{\sin }^2}\left( {{{90}^o} - \theta } \right)}}{{3\left( {{{\sec }^2}{{61}^0} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.{{\sec }^2}{{36}^o}}}{{2\left( {\cos e{c^2}{{65}^0} - {{\tan }^2}{{25}^o}} \right)}}\]
Answer
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Hint: First we have to know the trigonometric functions are real functions which relate an angle of a right-angle triangle to ratios of two side lengths. Then mention their trigonometric ratios. Using the trigonometric identities simplifying the given trigonometric expression.
Complete step by step answer:
Trigonometric functions also called circular functions, angle functions or goniometric functions. sine, cosine, tangent, cosecant, secant and cotangent are six trigonometric ratios. These six trigonometric ratios are abbreviated as \[\sin \], \[\cos \], \[\tan \], \[\csc \], \[\sec \], \[\cot \].
Given \[\dfrac{{{{\sin }^2}\theta + {{\sin }^2}\left( {{{90}^o} - \theta } \right)}}{{3\left( {{{\sec }^2}{{61}^0} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.{{\sec }^2}{{36}^o}}}{{2\left( {\cos e{c^2}{{65}^0} - {{\tan }^2}{{25}^o}} \right)}}\]---(1)
We know that \[\sec x = \dfrac{1}{{\cos x}}\]and \[\cos ecx = \dfrac{1}{{\sin x}}\], then the expression (1) becomes
\[ = \]\[\dfrac{{{{\sin }^2}\theta + {{\sin }^2}\left( {{{90}^o} - \theta } \right)}}{{3\left( {\dfrac{1}{{{{\cos }^2}{{61}^0}}} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.\dfrac{1}{{{{\cos }^2}{{36}^o}}}}}{{2\left( {\dfrac{1}{{{{\sin }^2}{{65}^o}}} - {{\tan }^2}{{25}^o}} \right)}}\]---(2)
We know that \[\sin \left( {{{90}^o} - \theta } \right) = \cos \theta \] then the expression (2) becomes
\[\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\left( \theta \right)}}{{3\left( {\dfrac{1}{{{{\cos }^2}{{61}^0}}} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.\dfrac{1}{{{{\cos }^2}{{36}^o}}}}}{{2\left( {\dfrac{1}{{{{\sin }^2}{{65}^o}}} - {{\tan }^2}{{25}^o}} \right)}}\]---(3)
Since \[{\sin ^2}x + {\cos ^2}x = 1\] then the expression (3) becomes
\[\dfrac{1}{{3\left( {\dfrac{1}{{{{\cos }^2}{{61}^0}}} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.\dfrac{1}{{{{\cos }^2}{{36}^o}}}}}{{2\left( {\dfrac{1}{{{{\sin }^2}{{65}^o}}} - {{\tan }^2}{{25}^o}} \right)}}\]------(4)
We know that \[\sin \left( {{{90}^o} - x} \right) = \cos x\]and \[\cos \left( {{{90}^o} - x} \right) = \sin x\].
Since \[\cos {61^o}\]can be expressed as \[\cos \left( {{{90}^o} - {{29}^o}} \right)\], since \[\cos \left( {{{90}^o} - {{29}^o}} \right) = \sin {29^o}\]then we get \[\cos {61^o} = \sin {29^o}\]. Also, \[\cos {36^o}\]can be expressed as \[\cos \left( {{{90}^o} - {{54}^o}} \right)\], since \[\cos \left( {{{90}^o} - {{54}^o}} \right) = \sin {54^o}\]then we get \[\cos {36^o} = \sin {54^o}\].
Similarly, \[\sin {65^o}\]can be expressed as \[\sin \left( {{{90}^o} - {{25}^o}} \right)\], since \[\sin \left( {{{90}^o} - {{25}^o}} \right) = \cos {25^o}\]then we get
\[\sin {65^o} = \cos {25^o}\].
Using the above formulas then the expression (4) becomes
\[\dfrac{1}{{3\left( {\dfrac{1}{{{{\sin }^2}{{29}^0}}} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.\dfrac{1}{{{{\sin }^2}{{54}^o}}}}}{{2\left( {\dfrac{1}{{{{\cos }^2}{{25}^o}}} - {{\tan }^2}{{25}^o}} \right)}}\]-----(5)
Again using \[\dfrac{1}{{\sin x}} = \cos ecx\]and \[\dfrac{1}{{\cos x}} = \sec x\]then the expression (6) becomes
\[\dfrac{1}{{3\left( {\cos e{c^2}{{29}^0} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}}}{{2\left( {{{\sec }^2}{{25}^o} - {{\tan }^2}{{25}^o}} \right)}}\]----(6)
We know that \[{\sec ^2}x - {\tan ^2}x = 1\]and \[\cos e{c^2}x - {\cot ^2}x = 1\], then the expression (6) becomes
\[\dfrac{1}{3} - \dfrac{{3{{\cot }^2}{{30}^o}}}{2}\]-----(7)
Since \[\cot {30^o} = \sqrt 3 \] then the expression (8) becomes
\[\dfrac{1}{3} - \dfrac{9}{2}\]
\[ \Rightarrow \]\[\dfrac{{2 - 27}}{6} = - \dfrac{{25}}{6}\].
Hence \[\dfrac{{{{\sin }^2}\theta + {{\sin }^2}\left( {{{90}^o} - \theta } \right)}}{{3\left( {{{\sec }^2}{{61}^0} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.{{\sec }^2}{{36}^o}}}{{2\left( {\cos e{c^2}{{65}^0} - {{\tan }^2}{{25}^o}} \right)}} = - \dfrac{{25}}{6}\].
Note: Note that to convert degree into radian we use the formula \[{1^o} = \dfrac{\pi }{{180}} \times radian\]. Similarly, to convert radian into degree we use the formula \[1\;radian = \dfrac{{180}}{\pi } \times \]degree. In geometry, trigonometric functions are used to find the unknown angle or side of a right-angled triangle.
Complete step by step answer:
Trigonometric functions also called circular functions, angle functions or goniometric functions. sine, cosine, tangent, cosecant, secant and cotangent are six trigonometric ratios. These six trigonometric ratios are abbreviated as \[\sin \], \[\cos \], \[\tan \], \[\csc \], \[\sec \], \[\cot \].
Given \[\dfrac{{{{\sin }^2}\theta + {{\sin }^2}\left( {{{90}^o} - \theta } \right)}}{{3\left( {{{\sec }^2}{{61}^0} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.{{\sec }^2}{{36}^o}}}{{2\left( {\cos e{c^2}{{65}^0} - {{\tan }^2}{{25}^o}} \right)}}\]---(1)
We know that \[\sec x = \dfrac{1}{{\cos x}}\]and \[\cos ecx = \dfrac{1}{{\sin x}}\], then the expression (1) becomes
\[ = \]\[\dfrac{{{{\sin }^2}\theta + {{\sin }^2}\left( {{{90}^o} - \theta } \right)}}{{3\left( {\dfrac{1}{{{{\cos }^2}{{61}^0}}} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.\dfrac{1}{{{{\cos }^2}{{36}^o}}}}}{{2\left( {\dfrac{1}{{{{\sin }^2}{{65}^o}}} - {{\tan }^2}{{25}^o}} \right)}}\]---(2)
We know that \[\sin \left( {{{90}^o} - \theta } \right) = \cos \theta \] then the expression (2) becomes
\[\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\left( \theta \right)}}{{3\left( {\dfrac{1}{{{{\cos }^2}{{61}^0}}} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.\dfrac{1}{{{{\cos }^2}{{36}^o}}}}}{{2\left( {\dfrac{1}{{{{\sin }^2}{{65}^o}}} - {{\tan }^2}{{25}^o}} \right)}}\]---(3)
Since \[{\sin ^2}x + {\cos ^2}x = 1\] then the expression (3) becomes
\[\dfrac{1}{{3\left( {\dfrac{1}{{{{\cos }^2}{{61}^0}}} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.\dfrac{1}{{{{\cos }^2}{{36}^o}}}}}{{2\left( {\dfrac{1}{{{{\sin }^2}{{65}^o}}} - {{\tan }^2}{{25}^o}} \right)}}\]------(4)
We know that \[\sin \left( {{{90}^o} - x} \right) = \cos x\]and \[\cos \left( {{{90}^o} - x} \right) = \sin x\].
Since \[\cos {61^o}\]can be expressed as \[\cos \left( {{{90}^o} - {{29}^o}} \right)\], since \[\cos \left( {{{90}^o} - {{29}^o}} \right) = \sin {29^o}\]then we get \[\cos {61^o} = \sin {29^o}\]. Also, \[\cos {36^o}\]can be expressed as \[\cos \left( {{{90}^o} - {{54}^o}} \right)\], since \[\cos \left( {{{90}^o} - {{54}^o}} \right) = \sin {54^o}\]then we get \[\cos {36^o} = \sin {54^o}\].
Similarly, \[\sin {65^o}\]can be expressed as \[\sin \left( {{{90}^o} - {{25}^o}} \right)\], since \[\sin \left( {{{90}^o} - {{25}^o}} \right) = \cos {25^o}\]then we get
\[\sin {65^o} = \cos {25^o}\].
Using the above formulas then the expression (4) becomes
\[\dfrac{1}{{3\left( {\dfrac{1}{{{{\sin }^2}{{29}^0}}} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.\dfrac{1}{{{{\sin }^2}{{54}^o}}}}}{{2\left( {\dfrac{1}{{{{\cos }^2}{{25}^o}}} - {{\tan }^2}{{25}^o}} \right)}}\]-----(5)
Again using \[\dfrac{1}{{\sin x}} = \cos ecx\]and \[\dfrac{1}{{\cos x}} = \sec x\]then the expression (6) becomes
\[\dfrac{1}{{3\left( {\cos e{c^2}{{29}^0} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}}}{{2\left( {{{\sec }^2}{{25}^o} - {{\tan }^2}{{25}^o}} \right)}}\]----(6)
We know that \[{\sec ^2}x - {\tan ^2}x = 1\]and \[\cos e{c^2}x - {\cot ^2}x = 1\], then the expression (6) becomes
\[\dfrac{1}{3} - \dfrac{{3{{\cot }^2}{{30}^o}}}{2}\]-----(7)
Since \[\cot {30^o} = \sqrt 3 \] then the expression (8) becomes
\[\dfrac{1}{3} - \dfrac{9}{2}\]
\[ \Rightarrow \]\[\dfrac{{2 - 27}}{6} = - \dfrac{{25}}{6}\].
Hence \[\dfrac{{{{\sin }^2}\theta + {{\sin }^2}\left( {{{90}^o} - \theta } \right)}}{{3\left( {{{\sec }^2}{{61}^0} - {{\cot }^2}{{29}^o}} \right)}} - \dfrac{{3{{\cot }^2}{{30}^o}.{{\sin }^2}{{54}^o}.{{\sec }^2}{{36}^o}}}{{2\left( {\cos e{c^2}{{65}^0} - {{\tan }^2}{{25}^o}} \right)}} = - \dfrac{{25}}{6}\].
Note: Note that to convert degree into radian we use the formula \[{1^o} = \dfrac{\pi }{{180}} \times radian\]. Similarly, to convert radian into degree we use the formula \[1\;radian = \dfrac{{180}}{\pi } \times \]degree. In geometry, trigonometric functions are used to find the unknown angle or side of a right-angled triangle.
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