Question

Check whether LHS is equal to RHS?
\[\sqrt {\dfrac{{{\text{cosec}}\theta - 1}}{{{\text{cosec}}\theta + 1}}} + \sqrt {\dfrac{{{\text{cosec}}\theta + 1}}{{{\text{cosec}}\theta - 1}}} = 2\cos \theta \] , Say true or false.
A. True
B. False
C. Ambiguous
D. Data insufficient

Answer 1

Hint: Here we will be solving this question by using the LCM (Least common factor) which is used for adding the fractions where denominators are not the same. It is the smallest positive integer that is a multiple of both the numbers of which we are taking LCM. For example, there are two fractions
\[\dfrac{a}{b}\] and \[\dfrac{c}{d}\] , then the addition of the terms by using LCM is as below:
\[\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{a + c}}{{bd}}\].
Also, we will be using the formula of \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\], \[a\] and \[b\] are two variables.

Complete step-by-step solution:
Step 1: We can write the given expression \[\sqrt {\dfrac{{{\text{cosec}}\theta - 1}}{{{\text{cosec}}\theta + 1}}} + \sqrt {\dfrac{{{\text{cosec}}\theta + 1}}{{{\text{cosec}}\theta - 1}}} = 2\cos \theta \] as below:
\[ \Rightarrow \dfrac{{\sqrt {{\text{cosec}}\theta - 1} }}{{\sqrt {{\text{cosec}}\theta + 1} }} + \dfrac{{\sqrt {{\text{cosec}}\theta + 1} }}{{\sqrt {{\text{cosec}}\theta - 1} }}\]
Step 2: Now by taking the LCM of the above expression we get:
\[ \Rightarrow \dfrac{{\left( {\sqrt {{\text{cosec}}\theta - 1} } \right)\left( {\sqrt {{\text{cosec}}\theta - 1} } \right) + \left( {\sqrt {{\text{cosec}}\theta + 1} } \right)\left( {\sqrt {{\text{cosec}}\theta + 1} } \right)}}{{\sqrt {\left( {{\text{cosec}}\theta + 1} \right)\left( {{\text{cosec}}\theta - 1} \right)} }}\]
We can write the above product of the same numbers as below:
\[ \Rightarrow \dfrac{{{{\left( {\sqrt {{\text{cosec}}\theta - 1} } \right)}^2} + {{\left( {\sqrt {{\text{cosec}}\theta + 1} } \right)}^2}}}{{\sqrt {\left( {{\text{cosec}}\theta + 1} \right)\left( {{\text{cosec}}\theta - 1} \right)} }}\] ,
\[\because a \times a = {a^2}\]
Step 3: Now by using the formula of \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\] in the denominator of the above expression we get:
\[ \Rightarrow \dfrac{{{{\left( {\sqrt {{\text{cosec}}\theta - 1} } \right)}^2} + {{\left( {\sqrt {{\text{cosec}}\theta + 1} } \right)}^2}}}{{\sqrt {{{\left( {{\text{cosec}}\theta } \right)}^2} - {1^2}} }}\]
By canceling the squares of the numerator terms with their square root we get:
\[ \Rightarrow \dfrac{{\left( {{\text{cosec}}\theta - 1} \right) + \left( {{\text{cosec}}\theta + 1} \right)}}{{\sqrt {{{\left( {{\text{cosec}}\theta } \right)}^2} - {1^2}} }}\] \[\because \left( {\sqrt {{a^2}} } \right) = a\]
By opening the brackets of numerator terms in the above expression we get:
\[ \Rightarrow \dfrac{{{\text{cosec}}\theta - 1 + {\text{cosec}}\theta + 1}}{{\sqrt {{{\left( {{\text{cosec}}\theta } \right)}^2} - {1^2}} }}\]
After doing the final addition and subtraction in the numerator of the above expression we get:
\[ \Rightarrow \dfrac{{{\text{2cosec}}\theta }}{{\sqrt {{{\left( {{\text{cosec}}\theta } \right)}^2} - {1^2}} }}\]
Step 4: As we know that \[{\cot ^2}\theta = {\text{cose}}{{\text{c}}^2}\theta - 1\], so by replacing it in the denominator of the above expression we get:
\[ \Rightarrow \dfrac{{{\text{2cosec}}\theta }}{{\sqrt {{{\cot }^2}\theta } }}\]
By canceling the square of the denominator of the above expression with the square root, we get:
\[ \Rightarrow \dfrac{{{\text{2cosec}}\theta }}{{\cot \theta }}\]
Step 5: As we know that \[{\text{cosec}}\theta {\text{ = }}\dfrac{1}{{\sin \theta }}\] and \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\], so by replacing these terms in the above expression we get:
\[ \Rightarrow \dfrac{{{\text{2sin}}\theta }}{{\sin \theta \cos \theta }}\]
After canceling \[\sin \theta \] from the above expression we get:
\[ \Rightarrow \dfrac{{\text{2}}}{{\cos \theta }}\]
By replacing \[\dfrac{1}{{\cos \theta }} = \sec \theta \] in the above expression we get:
\[ \Rightarrow 2\sec \theta \]
So, we can say that LHS does not equal RHS.

B is the correct option.

Note: Students should remember some of the terms mentioned below which plays an important role in solving these types of question:
\[\sin \theta = \dfrac{1}{{{\text{cosec}}\theta }}\]
\[\cos \theta = \dfrac{1}{{\sec \theta }}\]
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\tan \theta = \dfrac{1}{{\cot \theta }}\]