# Evaluate the value of: \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}\]

Last updated date: 25th Mar 2023

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Answer

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Hint: To evaluate the value of the given expression, substitute the value of given trigonometric functions at fixed angles and simplify them to get the exact value of the expression.

We have to evaluate the value of \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}\].

We will solve this by substituting the values of trigonometric functions at given angles.

We know that for \[\theta ={{30}^{\circ }}\], we have \[\tan \theta =\dfrac{1}{\sqrt{3}},\sin \theta =\dfrac{1}{2}\].

For \[\theta ={{45}^{\circ }}\], we have \[\tan \theta =1\].

For \[\theta ={{0}^{\circ }}\], we have \[\cos \theta =1\].

For \[\theta ={{60}^{\circ }}\], we have \[\cos \theta =\dfrac{1}{2},\tan \theta =\sqrt{3}\].

For \[\theta ={{90}^{\circ }}\], we have \[\sin \theta =1\].

Substituting the above values in the given expression, we have \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}={{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}\left( \dfrac{1}{2} \right)+\left( \dfrac{1}{2} \right){{\left( 1 \right)}^{2}}{{\left( \sqrt{3} \right)}^{2}}-2\left( 1 \right){{\left( 1 \right)}^{2}}\left( 1 \right)\]

Simplifying the above expression, we have \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}=\dfrac{1}{6}+\dfrac{3}{2}-2\].

Further simplifying, we get \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}=\dfrac{-1}{3}\].

Hence, the value of the given expression is \[\dfrac{-1}{3}\].

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius \[1\]). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one. One must keep in mind that the angles are given to us in degrees. If we solve them by considering them in radians and then converting them to degrees, we will get an incorrect answer.

We have to evaluate the value of \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}\].

We will solve this by substituting the values of trigonometric functions at given angles.

We know that for \[\theta ={{30}^{\circ }}\], we have \[\tan \theta =\dfrac{1}{\sqrt{3}},\sin \theta =\dfrac{1}{2}\].

For \[\theta ={{45}^{\circ }}\], we have \[\tan \theta =1\].

For \[\theta ={{0}^{\circ }}\], we have \[\cos \theta =1\].

For \[\theta ={{60}^{\circ }}\], we have \[\cos \theta =\dfrac{1}{2},\tan \theta =\sqrt{3}\].

For \[\theta ={{90}^{\circ }}\], we have \[\sin \theta =1\].

Substituting the above values in the given expression, we have \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}={{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}\left( \dfrac{1}{2} \right)+\left( \dfrac{1}{2} \right){{\left( 1 \right)}^{2}}{{\left( \sqrt{3} \right)}^{2}}-2\left( 1 \right){{\left( 1 \right)}^{2}}\left( 1 \right)\]

Simplifying the above expression, we have \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}=\dfrac{1}{6}+\dfrac{3}{2}-2\].

Further simplifying, we get \[{{\tan }^{2}}{{30}^{\circ }}\sin {{30}^{\circ }}+\cos {{60}^{\circ }}{{\sin }^{2}}{{90}^{\circ }}{{\tan }^{2}}{{60}^{\circ }}-2\tan {{45}^{\circ }}{{\cos }^{2}}{{0}^{\circ }}\sin {{90}^{\circ }}=\dfrac{-1}{3}\].

Hence, the value of the given expression is \[\dfrac{-1}{3}\].

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius \[1\]). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one. One must keep in mind that the angles are given to us in degrees. If we solve them by considering them in radians and then converting them to degrees, we will get an incorrect answer.

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