Question

If $\sin \theta + \cos ec\theta = 2$, then the value of ${\sin ^2}\theta + \cos e{c^2}\theta $ equals:
(A) $1$
(B) $4$
(C) $2$
(D) None of these

Answer 1

Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\cos ec(x) = \dfrac{1}{{\sin (x)}}$ and $\sin x = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$ . Basic algebraic rules such as ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and trigonometric identities are to be kept in mind while doing simplification in the given problem.

Complete step-by-step answer:
In the given problem, we are given the equation involving the trigonometric functions.
So, we have, $\sin \theta + \cos ec\theta = 2$.
Now, we square both the sides of the trigonometric equations so as to simplify it and get to our required expression. So, we get,
$ \Rightarrow {\left( {\sin \theta + \cos ec\theta } \right)^2} = {2^2}$
Now, we compute the whole square of the left side of the equation using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. So, we get,
$ \Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta + 2\sin \theta \cos ec\theta = 4$
Now, we know that the trigonometric functions sine and cosecant are reciprocal functions of each other. So, we get,
$ \Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta + 2\left( 1 \right) = 4$
Shifting all the constant to the right side of the equation, we get,
$ \Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta = 4 - 2$
Doing the calculations,
$ \Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta = 2$
Hence, the value of expression ${\sin ^2}\theta + \cos e{c^2}\theta $ is $2$ by the use of basic algebraic rules and simple trigonometric formulae.
So, the correct answer is “Option B”.

Note: Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers. The given problem can also be solved by first forming a quadratic equation in sine and solving it to get the value of sine trigonometric function and then using the value of sine to find the value of required expression.