Answer
Verified
412.2k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE