Answer
385.5k+ views
Hint: To solve the given question, we should know some of the trigonometric properties that are given below, we should know that sine and cosine are inverse of secant and cosecant function respectively, that is \[\sec \theta =\dfrac{1}{\cos \theta }\And \csc \theta =\dfrac{1}{\sin \theta }\]. Also, we should know the trigonometric identity relation between sine-cosine function, and secant-tangent functions,
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\And 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \].
Complete step by step answer:
The first statement we need to prove is, \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}=2cosec\theta \]. The LHS of the statement is \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}\], and the RHS is \[2cosec\theta \]. Multiplying the first term in the LHS by \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta -1}}\], and the second term by \[\sqrt{\dfrac{\sec \theta +1}{\sec \theta +1}}\]. We get
\[\Rightarrow \sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}\sqrt{\dfrac{\sec \theta -1}{\sec \theta -1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}\sqrt{\dfrac{\sec \theta +1}{\sec \theta +1}}\]
Simplifying the above expression, it can be written as
\[\Rightarrow \sqrt{\dfrac{{{\left( \sec \theta -1 \right)}^{2}}}{{{\left( \sec \theta \right)}^{2}}-{{\left( 1 \right)}^{2}}}}+\sqrt{\dfrac{{{\left( \sec \theta +1 \right)}^{2}}}{{{\left( \sec \theta \right)}^{2}}-{{\left( 1 \right)}^{2}}}}\]
Using the trigonometric identity, \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \], the denominator of the above expression can be expressed as
\[\Rightarrow \sqrt{\dfrac{{{\left( \sec \theta -1 \right)}^{2}}}{{{\tan }^{2}}\theta }}+\sqrt{\dfrac{{{\left( \sec \theta +1 \right)}^{2}}}{{{\tan }^{2}}\theta }}\]
canceling out the power, we get
\[\Rightarrow \dfrac{\left( \sec \theta -1 \right)}{\tan \theta }+\dfrac{\left( \sec \theta +1 \right)}{\tan \theta }\]
As the denominator of both terms is the same, we can add the numerator directly, by doing this we get
\[\Rightarrow \dfrac{2\sec \theta }{\tan \theta }\]
We know that \[\sec \theta =\dfrac{1}{\cos \theta }\And \tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Using this, the above expression can be simplified as
\[\Rightarrow \dfrac{2}{\sin \theta }=2\csc \theta =RHS\]
\[\therefore LHS=RHS\]
Hence, proved.
The second statement to prove is \[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}=2\sec \theta \]. The LHS of the statement is \[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}\], and the RHS of the statement is \[2\sec \theta \].
Multiplying the first term in the LHS by \[\sqrt{\dfrac{1+\sin \theta }{1+\sin \theta }}\], and second term by \[\sqrt{\dfrac{1-\sin \theta }{1-\sin \theta }}\]. We get
\[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}\sqrt{\dfrac{1+\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}\sqrt{\dfrac{1-\sin \theta }{1-\sin \theta }}\]
Simplifying the above expression, we get
\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{1}^{2}}-{{\sin }^{2}}\theta }}+\sqrt{\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{1}^{2}}-{{\sin }^{2}}\theta }}\]
Using the trigonometric identity, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], the denominator of the above expression can be expressed as
\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}+\sqrt{\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}\]
canceling out the square and square root from the above expression, we get
\[\Rightarrow \dfrac{\left( 1+\sin \theta \right)}{\cos \theta }+\dfrac{\left( 1-\sin \theta \right)}{\cos \theta }\]
As the denominator of both terms is the same, we can add the numerator directly, by doing this we get
\[\Rightarrow \dfrac{2}{\cos \theta }\]
Using \[\sec \theta =\dfrac{1}{\cos \theta }\], the above expression can be written as
\[\Rightarrow 2\sec \theta =RHS\]
\[\therefore LHS=RHS\]
Hence, proved.
Note: To solve problems based on trigonometric functions, one should remember the trigonometric properties and the identities. The properties we used to prove the given statements are \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\And 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \], also \[\sec \theta =\dfrac{1}{\cos \theta }\And \csc \theta =\dfrac{1}{\sin \theta }\].
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\And 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \].
Complete step by step answer:
The first statement we need to prove is, \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}=2cosec\theta \]. The LHS of the statement is \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}\], and the RHS is \[2cosec\theta \]. Multiplying the first term in the LHS by \[\sqrt{\dfrac{\sec \theta -1}{\sec \theta -1}}\], and the second term by \[\sqrt{\dfrac{\sec \theta +1}{\sec \theta +1}}\]. We get
\[\Rightarrow \sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}\sqrt{\dfrac{\sec \theta -1}{\sec \theta -1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}\sqrt{\dfrac{\sec \theta +1}{\sec \theta +1}}\]
Simplifying the above expression, it can be written as
\[\Rightarrow \sqrt{\dfrac{{{\left( \sec \theta -1 \right)}^{2}}}{{{\left( \sec \theta \right)}^{2}}-{{\left( 1 \right)}^{2}}}}+\sqrt{\dfrac{{{\left( \sec \theta +1 \right)}^{2}}}{{{\left( \sec \theta \right)}^{2}}-{{\left( 1 \right)}^{2}}}}\]
Using the trigonometric identity, \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \], the denominator of the above expression can be expressed as
\[\Rightarrow \sqrt{\dfrac{{{\left( \sec \theta -1 \right)}^{2}}}{{{\tan }^{2}}\theta }}+\sqrt{\dfrac{{{\left( \sec \theta +1 \right)}^{2}}}{{{\tan }^{2}}\theta }}\]
canceling out the power, we get
\[\Rightarrow \dfrac{\left( \sec \theta -1 \right)}{\tan \theta }+\dfrac{\left( \sec \theta +1 \right)}{\tan \theta }\]
As the denominator of both terms is the same, we can add the numerator directly, by doing this we get
\[\Rightarrow \dfrac{2\sec \theta }{\tan \theta }\]
We know that \[\sec \theta =\dfrac{1}{\cos \theta }\And \tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Using this, the above expression can be simplified as
\[\Rightarrow \dfrac{2}{\sin \theta }=2\csc \theta =RHS\]
\[\therefore LHS=RHS\]
Hence, proved.
The second statement to prove is \[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}=2\sec \theta \]. The LHS of the statement is \[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}\], and the RHS of the statement is \[2\sec \theta \].
Multiplying the first term in the LHS by \[\sqrt{\dfrac{1+\sin \theta }{1+\sin \theta }}\], and second term by \[\sqrt{\dfrac{1-\sin \theta }{1-\sin \theta }}\]. We get
\[\sqrt{\dfrac{1+\sin \theta }{1-\sin \theta }}\sqrt{\dfrac{1+\sin \theta }{1+\sin \theta }}+\sqrt{\dfrac{1-\sin \theta }{1+\sin \theta }}\sqrt{\dfrac{1-\sin \theta }{1-\sin \theta }}\]
Simplifying the above expression, we get
\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{1}^{2}}-{{\sin }^{2}}\theta }}+\sqrt{\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{1}^{2}}-{{\sin }^{2}}\theta }}\]
Using the trigonometric identity, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], the denominator of the above expression can be expressed as
\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}+\sqrt{\dfrac{{{\left( 1-\sin \theta \right)}^{2}}}{{{\cos }^{2}}\theta }}\]
canceling out the square and square root from the above expression, we get
\[\Rightarrow \dfrac{\left( 1+\sin \theta \right)}{\cos \theta }+\dfrac{\left( 1-\sin \theta \right)}{\cos \theta }\]
As the denominator of both terms is the same, we can add the numerator directly, by doing this we get
\[\Rightarrow \dfrac{2}{\cos \theta }\]
Using \[\sec \theta =\dfrac{1}{\cos \theta }\], the above expression can be written as
\[\Rightarrow 2\sec \theta =RHS\]
\[\therefore LHS=RHS\]
Hence, proved.
Note: To solve problems based on trigonometric functions, one should remember the trigonometric properties and the identities. The properties we used to prove the given statements are \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\And 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \], also \[\sec \theta =\dfrac{1}{\cos \theta }\And \csc \theta =\dfrac{1}{\sin \theta }\].
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)