Question

If \[\sin \theta + \csc \theta = 2\], then find the value of \[{\sin ^2}\theta + {\csc ^2}\theta \].
A. \[1\]
B. \[4\]
C. \[2\]
D. None of these

Answer 1

Hint:
If you take \[a = \sin \theta \] and \[b = \csc \theta \], then the question is like the value of \[a + b\] is given and the value of \[{a^2} + {b^2}\] is required. So, simply use the identity \[{a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab\], put the value of \[a + b\] i.e. \[\sin \theta + \csc \theta \] and the product is \[ab = \sin \theta \csc \theta = 1\]. After that putting these values and simplifying you’ll get the value of \[{a^2} + {b^2} = \,{\sin ^2}\theta + {\csc ^2}\theta \].



Formula Used:
\[{a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab\]
\[\sin \theta \csc \theta = 1\]


Complete step-by-step answer:
Given that \[\sin \theta + \csc \theta = 2 - - - - - \left( i \right)\]
Use the identity \[{a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab\]
Put \[a = \sin \theta \] and \[b = \csc \theta \] in the identity.
\[{\sin ^2}\theta + {\csc ^2}\theta = {\left( {\sin \theta + \csc \theta } \right)^2} - 2\sin \theta \csc \theta \]
Substitute \[\sin \theta + \csc \theta = 2\] from equation \[\left( i \right)\] and \[\sin \theta \csc \theta = 1\]
So, \[{\sin ^2}\theta + {\csc ^2}\theta = {\left( 2 \right)^2} - 2 \times 1 \]
\[{\sin ^2}\theta + {\csc ^2}\theta = 4 - 2 \]
\[{\sin ^2}\theta + {\csc ^2}\theta = 2\]
Hence option C is correct.



Note:
There are many other methods for solving this problem.
One of the methods is shown below.
Squaring the given equation, we get
\[{\left( {\sin \theta + \csc \theta } \right)^2} = 4\]
Use the identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[ \Rightarrow \,{\sin ^2}\theta + {\csc ^2}\theta + 2\sin \theta \csc \theta = 4\]
Use the identity \[\sin \theta \csc \theta = 1\]
\[ \Rightarrow \,{\sin ^2}\theta + {\csc ^2}\theta + 2 = 4\]
\[ \Rightarrow \,{\sin ^2}\theta + {\csc ^2}\theta = 4 - 2 = 2\]