Question

The value of \[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ \] is equal to
(A) \[{{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ \]
(B) \[{{\tan }^{2}}45{}^\circ +{{\cot }^{2}}45{}^\circ \]
(C) \[{{\sec }^{2}}90{}^\circ \]
(D) \[0\]

Answer 1

Hint: It is given that our expression is \[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ \] . We know that, \[\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}\] and \[\cos 60{}^\circ =\dfrac{1}{2}\] . Now, put these values in the expression, \[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ \] and calculate its value. We also know that, \[\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}\] and \[\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}\] . Now, put these values in the expression \[{{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ \] and solve it further. Now, compare values and conclude the answer.

Complete step by step solution:
According to the question, it is given that our expression is
\[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ \] …………………..(1)
In this equation, we have summation of two terms \[{{\sin }^{2}}60{}^\circ \] and \[{{\cos }^{2}}60{}^\circ \] . To solve this equation, we need the value of \[\sin 60{}^\circ \] and \[\cos 60{}^\circ \] .
We know that, \[\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}\] …………………….(2)
Now, squaring LHS and RHS of equation (2), we get
 \[{{\sin }^{2}}60{}^\circ ={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}\]
\[\Rightarrow {{\sin }^{2}}60{}^\circ =\dfrac{3}{4}\] ………………….(3)
We know that, \[\cos 60{}^\circ =\dfrac{1}{2}\] …………………….(4)
Now, squaring LHS and RHS of equation (4), we get
 \[{{\cos }^{2}}60{}^\circ ={{\left( \dfrac{1}{2} \right)}^{2}}\]
\[\Rightarrow {{\cos }^{2}}60{}^\circ =\dfrac{1}{4}\] ………………….(5)
From, equation (1), we have \[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ \] and from equation (3) and equation (5), we have the value of \[{{\sin }^{2}}60{}^\circ \] and \[{{\cos }^{2}}60{}^\circ \] .
Now, putting the value of \[{{\sin }^{2}}60{}^\circ \] and \[{{\cos }^{2}}60{}^\circ \] in equation (1), we get
\[\begin{align}
  & {{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ \\
 & =\dfrac{3}{4}+\dfrac{1}{4} \\
 & =\dfrac{4}{4} \\
 & =1 \\
\end{align}\]
So, \[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ =1\] …………………(6)
Now, we have to get the values of the following options by solving them.
In option (A), we have \[{{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ \] ……………..(7)
We know that, \[\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}\] …………………….(8)
Now, squaring LHS and RHS of equation (8), we get
 \[{{\sin }^{2}}45{}^\circ ={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}\]
\[\Rightarrow {{\sin }^{2}}45{}^\circ =\dfrac{1}{2}\] ………………….(9)
We know that, \[\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}\] …………………….(10)
Now, squaring LHS and RHS of equation (10), we get
 \[{{\cos }^{2}}45{}^\circ ={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}\]
\[\Rightarrow {{\cos }^{2}}45{}^\circ =\dfrac{1}{2}\] ………………….(11)
From, equation (7), we have \[{{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ \] and from equation (9) and equation (11), we have the value of \[{{\sin }^{2}}45{}^\circ \] and \[{{\cos }^{2}}45{}^\circ \] .
Now, putting the value of \[{{\sin }^{2}}45{}^\circ \] and \[{{\cos }^{2}}45{}^\circ \] in equation (7), we get
\[\begin{align}
  & {{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ \\
 & =\dfrac{1}{2}+\dfrac{1}{2} \\
 & =\dfrac{2}{2} \\
 & =1 \\
\end{align}\]
So, \[{{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ =1\] …………………(12)
From equation (6), we have \[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ =1\] .
From equation (12), we have \[{{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ =1\] .
The RHS of equation (6) and equation (12) is equal to 1. So, the LHS of equation (6) and equation (12) must also be equal.
Therefore, \[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ ={{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ \] .
Hence, the correct option is (A).

Note: We can also solve this question without using the value of \[\sin 60{}^\circ \] and \[\cos 60{}^\circ \] .
We know the identity, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] ……………….(1)
Now, replacing \[\theta \] by \[60{}^\circ \] in equation (1), we get
\[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ =1\] …………………..(2)
Replacing \[\theta \] by \[45{}^\circ \] in equation (1), we get
\[{{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ =1\] …………………….(3)
The RHS of equation (2) and equation (3) is equal to 1. So, the LHS of equation (2) and equation (3) must also be equal.
Therefore, \[{{\sin }^{2}}60{}^\circ +{{\cos }^{2}}60{}^\circ ={{\sin }^{2}}45{}^\circ +{{\cos }^{2}}45{}^\circ \] .