Evaluate the following trigonometric expression using trigonometric identities: $ \dfrac{2\sin {{68}^{\circ }}}{\cos {{22}^{\circ }}}-\dfrac{2\cot {{15}^{\circ }}}{5\tan {{75}^{\circ }}}-\dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5} $ \[\]
Answer
Verified574.5k+ views
Hint: We convert the sine into cosine in the first term using the reduction formula $ \sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta $ and covert the tangents and cotangents using the reduction formula in the second and third term as $ \tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ and the reciprocal relation $ \tan \theta =\dfrac{1}{\cot \theta } $ where $ \theta $ is the smaller angle. \[\]
Complete step-by-step answer:
We know that when sum measures of two angles are equal to the measure of right angles that is $ {{90}^{\circ }} $ we call them complementary angles. Let us assume $ \alpha ,\beta $ be the measure of two complementary angles then we have,
\[\alpha +\beta ={{90}^{\circ }}\]
If we assume that $ \alpha =\theta $ then we can express both the angles in terms of $ \theta $ as,
\[\alpha =\theta ,\beta ={{90}^{\circ }}-\theta \]
We also know about the six trigonometric functions sine, cosine, tangent, cotangent, secant and cosecant defined on angle $ \theta $ as $ \sin \theta ,\cos \theta ,\tan \theta ,\cot \theta ,\sec \theta ,\operatorname{cosec}\theta $ respectively. The pair of functions sine and cosine, tangent and cotangent, secant and cosecant are called complementary trigonometric function because they obey relations for two complementary angles $ \theta $ and $ {{90}^{\circ }}-\theta $ as given below,
\[\begin{align}
& \sin \theta =\cos \left( {{90}^{\circ }}-\theta \right),\tan \theta =\cot \left( {{90}^{\circ }}-\theta \right),\sec \theta =\operatorname{cosec}\left( {{90}^{\circ }}-\theta \right) \\
& \cos \theta =\sin \left( {{90}^{\circ }}-\theta \right),\cot \theta =\tan \left( {{90}^{\circ }}-\theta \right),\operatorname{cosec}\theta =\sec \left( {{90}^{\circ }}-\theta \right) \\
\end{align}\]
The above relations are also called reduction formulas. We know that there is a reciprocal relation between sine and cosec, cosine and sec, tangent and cotangent as given below.
\[\begin{align}
& \sin \theta =\dfrac{1}{\operatorname{cosec}\theta },\cos \theta =\dfrac{1}{\sec \theta },\tan \theta =\dfrac{1}{\cot \theta } \\
& \operatorname{cosec}\theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
We are given an expression in trigonometric functions to evaluate as follows,
\[\dfrac{2\sin {{68}^{\circ }}}{\cos {{22}^{\circ }}}-\dfrac{2\cot {{15}^{\circ }}}{5\tan {{75}^{\circ }}}-\dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5}......\left( 1 \right)\]
We see that we are the trigonometric functions of sine and cosine in the first term, tangent and cotangent in both the second and third term of the expression.
We observe the angles in the first term $ {{68}^{\circ }},{{22}^{\circ }} $ and add them to get $ {{68}^{\circ }}+{{22}^{\circ }}={{90}^{\circ }} $ . So they are complementary angles. We assume the smaller angle $ \theta ={{22}^{\circ }} $ and have $ {{90}^{\circ }}-\theta ={{68}^{\circ }} $ .We use the reduction formula for sine and cosine for $ \theta ={{22}^{\circ }} $ and have,
\[\dfrac{2\sin {{68}^{\circ }}}{\cos {{22}^{\circ }}}=\dfrac{2\sin \left( {{90}^{\circ }}-{{22}^{\circ }} \right)}{\cos {{22}^{\circ }}}=\dfrac{2\cos {{22}^{\circ }}}{\cos {{22}^{\circ }}}=2.....\left( 2 \right)\]
We observe the second term in the expression which has angles $ {{15}^{\circ }},{{75}^{\circ }} $ and add them to get $ {{15}^{\circ }}+{{75}^{\circ }}={{90}^{\circ }} $ . If we assume the smaller angle as $ \theta ={{15}^{\circ }} $ then we have $ {{90}^{\circ }}-\theta ={{75}^{\circ }} $ . We use the reduction formula for tangent and cotangent for \[\theta ={{15}^{\circ }}\] and have,
\[\dfrac{2\cot {{15}^{\circ }}}{5\tan {{75}^{\circ }}}=\dfrac{2\cot {{15}^{\circ }}}{5\tan \left( {{90}^{\circ }}-{{15}^{\circ }} \right)}=\dfrac{2\cot {{15}^{\circ }}}{5\cot {{15}^{\circ }}}=\dfrac{2}{5}.......\left( 3 \right)\]
We observe the third term in the expression which has angles $ {{45}^{\circ }},{{20}^{\circ }},{{40}^{\circ }},{{50}^{\circ }},{{70}^{\circ }} $ . We know that $ \tan {{45}^{\circ }}=1 $ . We also observe that if we add $ {{20}^{\circ }},{{70}^{\circ }} $ we get $ {{20}^{\circ }}+{{70}^{\circ }}={{90}^{\circ }} $ and also if we add $ {{40}^{\circ }},{{50}^{\circ }} $ we get $ {{40}^{\circ }}+{{50}^{\circ }}={{90}^{\circ }} $ . So they are a pair of complementary angles. If we take $ \theta ={{20}^{\circ }} $ we have $ {{90}^{\circ }}-\theta ={{70}^{\circ }} $ and if we take $ \theta ={{40}^{\circ }} $ we have $ {{90}^{\circ }}-\theta ={{50}^{\circ }} $ .We use the reduction formula for tangent and cotangent for $ \theta ={{20}^{\circ }},{{40}^{\circ }} $ and have
\[ \begin{align}
& \dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5} \\
& =\dfrac{3\times 1\times \tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan \left( {{90}^{\circ }}-{{40}^{\circ }} \right)\tan \left( {{90}^{\circ }}-{{20}^{\circ }} \right)}{5} \\
& =\dfrac{3\tan {{20}^{\circ }}\tan {{40}^{\circ }}\cot {{40}^{\circ }}\cot {{20}^{\circ }}}{5} \\
& =\dfrac{3\tan {{20}^{\circ }}\cot {{20}^{\circ }}\tan {{40}^{\circ }}\cot {{40}^{\circ }}}{5} \\
\end{align}\]
We use the reciprocal relationship between tangent and cotangent for $ \theta ={{20}^{\circ }},{{40}^{\circ }} $
\[\begin{align}
& \Rightarrow \dfrac{3\times \dfrac{1}{\cot {{20}^{\circ }}}\times \cot {{20}^{\circ }}\times \dfrac{1}{\cot {{40}^{\circ }}}\times \cot {{40}^{\circ }}}{5} \\
& \Rightarrow \dfrac{3\times 1\times 1}{5}=\dfrac{3}{5} \\
& \therefore \dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5}=\dfrac{3}{5}.....\left( 4 \right) \\
\end{align}\]
We put the obtained simplified values from equation (2), (3) and (4) in equation (1) to have,
\[\begin{align}
& \Rightarrow \dfrac{2\sin {{68}^{\circ }}}{\cos {{22}^{\circ }}}-\dfrac{2\cot {{15}^{\circ }}}{5\tan {{75}^{\circ }}}-\dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5} \\
& \Rightarrow 2-\dfrac{2}{5}-\dfrac{3}{5} \\
& \Rightarrow 2-\left( \dfrac{2+3}{5} \right) \\
& \Rightarrow 2-\dfrac{5}{5}=2-1=1 \\
\end{align} \]
So the evaluated value obtained is 1. \[\]
Note: We have assumed $ \theta $ as the smaller angle but we can take the greater angle as also $ \theta $ . We then have to convert cosine into sine $ \left( \cos \left( 90-\theta \right)=\sin \theta \right) $ and cotangent into tangent $ \cot \left( 90-\theta \right)=\tan \theta $ . The reduction formula is also called reflection identity because $ \theta $ and $ {{90}^{\circ }}-\theta $ reflect each other about the angle vector $ \alpha =\dfrac{\pi }{4} $ .
Complete step-by-step answer:
We know that when sum measures of two angles are equal to the measure of right angles that is $ {{90}^{\circ }} $ we call them complementary angles. Let us assume $ \alpha ,\beta $ be the measure of two complementary angles then we have,
\[\alpha +\beta ={{90}^{\circ }}\]
If we assume that $ \alpha =\theta $ then we can express both the angles in terms of $ \theta $ as,
\[\alpha =\theta ,\beta ={{90}^{\circ }}-\theta \]
We also know about the six trigonometric functions sine, cosine, tangent, cotangent, secant and cosecant defined on angle $ \theta $ as $ \sin \theta ,\cos \theta ,\tan \theta ,\cot \theta ,\sec \theta ,\operatorname{cosec}\theta $ respectively. The pair of functions sine and cosine, tangent and cotangent, secant and cosecant are called complementary trigonometric function because they obey relations for two complementary angles $ \theta $ and $ {{90}^{\circ }}-\theta $ as given below,
\[\begin{align}
& \sin \theta =\cos \left( {{90}^{\circ }}-\theta \right),\tan \theta =\cot \left( {{90}^{\circ }}-\theta \right),\sec \theta =\operatorname{cosec}\left( {{90}^{\circ }}-\theta \right) \\
& \cos \theta =\sin \left( {{90}^{\circ }}-\theta \right),\cot \theta =\tan \left( {{90}^{\circ }}-\theta \right),\operatorname{cosec}\theta =\sec \left( {{90}^{\circ }}-\theta \right) \\
\end{align}\]
The above relations are also called reduction formulas. We know that there is a reciprocal relation between sine and cosec, cosine and sec, tangent and cotangent as given below.
\[\begin{align}
& \sin \theta =\dfrac{1}{\operatorname{cosec}\theta },\cos \theta =\dfrac{1}{\sec \theta },\tan \theta =\dfrac{1}{\cot \theta } \\
& \operatorname{cosec}\theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
We are given an expression in trigonometric functions to evaluate as follows,
\[\dfrac{2\sin {{68}^{\circ }}}{\cos {{22}^{\circ }}}-\dfrac{2\cot {{15}^{\circ }}}{5\tan {{75}^{\circ }}}-\dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5}......\left( 1 \right)\]
We see that we are the trigonometric functions of sine and cosine in the first term, tangent and cotangent in both the second and third term of the expression.
We observe the angles in the first term $ {{68}^{\circ }},{{22}^{\circ }} $ and add them to get $ {{68}^{\circ }}+{{22}^{\circ }}={{90}^{\circ }} $ . So they are complementary angles. We assume the smaller angle $ \theta ={{22}^{\circ }} $ and have $ {{90}^{\circ }}-\theta ={{68}^{\circ }} $ .We use the reduction formula for sine and cosine for $ \theta ={{22}^{\circ }} $ and have,
\[\dfrac{2\sin {{68}^{\circ }}}{\cos {{22}^{\circ }}}=\dfrac{2\sin \left( {{90}^{\circ }}-{{22}^{\circ }} \right)}{\cos {{22}^{\circ }}}=\dfrac{2\cos {{22}^{\circ }}}{\cos {{22}^{\circ }}}=2.....\left( 2 \right)\]
We observe the second term in the expression which has angles $ {{15}^{\circ }},{{75}^{\circ }} $ and add them to get $ {{15}^{\circ }}+{{75}^{\circ }}={{90}^{\circ }} $ . If we assume the smaller angle as $ \theta ={{15}^{\circ }} $ then we have $ {{90}^{\circ }}-\theta ={{75}^{\circ }} $ . We use the reduction formula for tangent and cotangent for \[\theta ={{15}^{\circ }}\] and have,
\[\dfrac{2\cot {{15}^{\circ }}}{5\tan {{75}^{\circ }}}=\dfrac{2\cot {{15}^{\circ }}}{5\tan \left( {{90}^{\circ }}-{{15}^{\circ }} \right)}=\dfrac{2\cot {{15}^{\circ }}}{5\cot {{15}^{\circ }}}=\dfrac{2}{5}.......\left( 3 \right)\]
We observe the third term in the expression which has angles $ {{45}^{\circ }},{{20}^{\circ }},{{40}^{\circ }},{{50}^{\circ }},{{70}^{\circ }} $ . We know that $ \tan {{45}^{\circ }}=1 $ . We also observe that if we add $ {{20}^{\circ }},{{70}^{\circ }} $ we get $ {{20}^{\circ }}+{{70}^{\circ }}={{90}^{\circ }} $ and also if we add $ {{40}^{\circ }},{{50}^{\circ }} $ we get $ {{40}^{\circ }}+{{50}^{\circ }}={{90}^{\circ }} $ . So they are a pair of complementary angles. If we take $ \theta ={{20}^{\circ }} $ we have $ {{90}^{\circ }}-\theta ={{70}^{\circ }} $ and if we take $ \theta ={{40}^{\circ }} $ we have $ {{90}^{\circ }}-\theta ={{50}^{\circ }} $ .We use the reduction formula for tangent and cotangent for $ \theta ={{20}^{\circ }},{{40}^{\circ }} $ and have
\[ \begin{align}
& \dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5} \\
& =\dfrac{3\times 1\times \tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan \left( {{90}^{\circ }}-{{40}^{\circ }} \right)\tan \left( {{90}^{\circ }}-{{20}^{\circ }} \right)}{5} \\
& =\dfrac{3\tan {{20}^{\circ }}\tan {{40}^{\circ }}\cot {{40}^{\circ }}\cot {{20}^{\circ }}}{5} \\
& =\dfrac{3\tan {{20}^{\circ }}\cot {{20}^{\circ }}\tan {{40}^{\circ }}\cot {{40}^{\circ }}}{5} \\
\end{align}\]
We use the reciprocal relationship between tangent and cotangent for $ \theta ={{20}^{\circ }},{{40}^{\circ }} $
\[\begin{align}
& \Rightarrow \dfrac{3\times \dfrac{1}{\cot {{20}^{\circ }}}\times \cot {{20}^{\circ }}\times \dfrac{1}{\cot {{40}^{\circ }}}\times \cot {{40}^{\circ }}}{5} \\
& \Rightarrow \dfrac{3\times 1\times 1}{5}=\dfrac{3}{5} \\
& \therefore \dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5}=\dfrac{3}{5}.....\left( 4 \right) \\
\end{align}\]
We put the obtained simplified values from equation (2), (3) and (4) in equation (1) to have,
\[\begin{align}
& \Rightarrow \dfrac{2\sin {{68}^{\circ }}}{\cos {{22}^{\circ }}}-\dfrac{2\cot {{15}^{\circ }}}{5\tan {{75}^{\circ }}}-\dfrac{3\tan {{45}^{\circ }}\tan {{20}^{\circ }}\tan {{40}^{\circ }}\tan {{50}^{\circ }}\tan {{70}^{\circ }}}{5} \\
& \Rightarrow 2-\dfrac{2}{5}-\dfrac{3}{5} \\
& \Rightarrow 2-\left( \dfrac{2+3}{5} \right) \\
& \Rightarrow 2-\dfrac{5}{5}=2-1=1 \\
\end{align} \]
So the evaluated value obtained is 1. \[\]
Note: We have assumed $ \theta $ as the smaller angle but we can take the greater angle as also $ \theta $ . We then have to convert cosine into sine $ \left( \cos \left( 90-\theta \right)=\sin \theta \right) $ and cotangent into tangent $ \cot \left( 90-\theta \right)=\tan \theta $ . The reduction formula is also called reflection identity because $ \theta $ and $ {{90}^{\circ }}-\theta $ reflect each other about the angle vector $ \alpha =\dfrac{\pi }{4} $ .
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 English: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Biology: Engaging Questions & Answers for Success
Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE
Discuss the various forms of bacteria class 11 biology CBSE
Draw a diagram of a plant cell and label at least eight class 11 biology CBSE
State the laws of reflection of light
Explain zero factorial class 11 maths CBSE