# Evaluate the following: \[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}\].

Last updated date: 22nd Mar 2023

•

Total views: 308.4k

•

Views today: 6.86k

Answer

Verified

308.4k+ views

Hint: Transform the whole equation in terms of \[\sin \theta \]and \[\cos \theta \].

We have to evaluate: \[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}...\left( i \right)\]

We know that \[\sin \theta =\dfrac{1}{\text{cosec}\theta }\]and \[\cos \theta =\dfrac{1}{\text{sec}\theta

}\].

Therefore, \[\sec {{50}^{o}}=\dfrac{1}{\cos {{50}^{o}}}\]and

\[\operatorname{cosec}{{50}^{o}}=\dfrac{1}{\sin {{50}^{o}}}\]

Now, we will put these values in equation \[\left( i \right)\].

\[=\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\operatorname{cosec}{{50}^{o}}\]

We get, \[\dfrac{\sin {{40}^{o}}}{\cos {{50}^{o}}}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}....\left( ii \right)\]

Now, we know that \[\sin \left( {{90}^{o}}-\theta \right)=\cos \theta \]

For \[\theta ={{50}^{o}}\]

We get \[\sin \left( 90-{{50}^{o}} \right)=\cos {{50}^{o}}\]

\[\Rightarrow \sin {{40}^{o}}=\cos {{50}^{o}}\]

For \[\theta ={{40}^{o}}\]

We get \[\sin \left( 90-{{40}^{o}} \right)=\cos {{40}^{o}}\]

\[\Rightarrow \sin {{50}^{o}}=\cos {{40}^{o}}\]

Putting the values of \[\sin {{40}^{o}}\]and \[\cos {{50}^{o}}\]in equation \[\left( ii \right)\].

We get, \[\dfrac{\cos {{50}^{o}}}{\cos {{50}^{o}}}+\dfrac{\cos {{40}^{o}}}{\cos {{40}^{o}}}\]

By cancelling the terms, we get

\[=1+1\]

\[=2\]

Hence, we get \[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}=2\].

Note: This question can also be easily solved by converting \[\sec \left( 90-\theta \right)\]to

\[\operatorname{cosec}\theta \]and \[\operatorname{cosec}\left( 90-\theta \right)\]to\[\sec \theta \]as follows:

\[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}....\left( i \right)\]

We know that \[\sec \left( {{90}^{o}}-\theta \right)=\operatorname{cosec}\theta \]

For \[\theta ={{40}^{o}}\], we get \[\sec \left( {{90}^{o}}-{{40}^{o}}

\right)=\operatorname{cosec}{{40}^{o}}\]

Therefore, \[\sec {{50}^{o}}=\operatorname{cosec}{{40}^{o}}\]

Also, we know that \[\operatorname{cosec}\left( {{90}^{o}}-\theta \right)=\sec \theta \]

For \[\theta ={{40}^{o}}\], we get \[\operatorname{cosec}\left( {{90}^{o}}-{{40}^{o}}

\right)=sec{{40}^{o}}\]

Therefore, \[\operatorname{cosec}{{50}^{o}}=sec{{40}^{o}}\]

Now, we put values of \[\sec \left( {{50}^{o}} \right)\]and \[\operatorname{cosec}\left( {{50}^{o}}

\right)\]in equation \[\left( i \right)\].

We get, \[\operatorname{cosec}{{40}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\sec {{40}^{o}}\]

We know that \[\operatorname{cosec}\theta .\sin \theta =1\]and \[\cos \theta .\sec \theta =1\]

By putting it in above equation,

We get, \[1+1=2\]which is our required answer.

We have to evaluate: \[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}...\left( i \right)\]

We know that \[\sin \theta =\dfrac{1}{\text{cosec}\theta }\]and \[\cos \theta =\dfrac{1}{\text{sec}\theta

}\].

Therefore, \[\sec {{50}^{o}}=\dfrac{1}{\cos {{50}^{o}}}\]and

\[\operatorname{cosec}{{50}^{o}}=\dfrac{1}{\sin {{50}^{o}}}\]

Now, we will put these values in equation \[\left( i \right)\].

\[=\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\operatorname{cosec}{{50}^{o}}\]

We get, \[\dfrac{\sin {{40}^{o}}}{\cos {{50}^{o}}}+\dfrac{\cos {{40}^{o}}}{\sin {{50}^{o}}}....\left( ii \right)\]

Now, we know that \[\sin \left( {{90}^{o}}-\theta \right)=\cos \theta \]

For \[\theta ={{50}^{o}}\]

We get \[\sin \left( 90-{{50}^{o}} \right)=\cos {{50}^{o}}\]

\[\Rightarrow \sin {{40}^{o}}=\cos {{50}^{o}}\]

For \[\theta ={{40}^{o}}\]

We get \[\sin \left( 90-{{40}^{o}} \right)=\cos {{40}^{o}}\]

\[\Rightarrow \sin {{50}^{o}}=\cos {{40}^{o}}\]

Putting the values of \[\sin {{40}^{o}}\]and \[\cos {{50}^{o}}\]in equation \[\left( ii \right)\].

We get, \[\dfrac{\cos {{50}^{o}}}{\cos {{50}^{o}}}+\dfrac{\cos {{40}^{o}}}{\cos {{40}^{o}}}\]

By cancelling the terms, we get

\[=1+1\]

\[=2\]

Hence, we get \[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}=2\].

Note: This question can also be easily solved by converting \[\sec \left( 90-\theta \right)\]to

\[\operatorname{cosec}\theta \]and \[\operatorname{cosec}\left( 90-\theta \right)\]to\[\sec \theta \]as follows:

\[\sec {{50}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\text{cosec}{{50}^{o}}....\left( i \right)\]

We know that \[\sec \left( {{90}^{o}}-\theta \right)=\operatorname{cosec}\theta \]

For \[\theta ={{40}^{o}}\], we get \[\sec \left( {{90}^{o}}-{{40}^{o}}

\right)=\operatorname{cosec}{{40}^{o}}\]

Therefore, \[\sec {{50}^{o}}=\operatorname{cosec}{{40}^{o}}\]

Also, we know that \[\operatorname{cosec}\left( {{90}^{o}}-\theta \right)=\sec \theta \]

For \[\theta ={{40}^{o}}\], we get \[\operatorname{cosec}\left( {{90}^{o}}-{{40}^{o}}

\right)=sec{{40}^{o}}\]

Therefore, \[\operatorname{cosec}{{50}^{o}}=sec{{40}^{o}}\]

Now, we put values of \[\sec \left( {{50}^{o}} \right)\]and \[\operatorname{cosec}\left( {{50}^{o}}

\right)\]in equation \[\left( i \right)\].

We get, \[\operatorname{cosec}{{40}^{o}}\sin {{40}^{o}}+\cos {{40}^{o}}\sec {{40}^{o}}\]

We know that \[\operatorname{cosec}\theta .\sin \theta =1\]and \[\cos \theta .\sec \theta =1\]

By putting it in above equation,

We get, \[1+1=2\]which is our required answer.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

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

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE