Question

Find the value of ${\cos ^2}{60^ \circ }{\sin ^2}{30^ \circ } + {\tan ^2}{30^ \circ }{\cot ^2}{60^ \circ }$ ?

Answer 1

Hint: We will solve the trigonometric expression ${\cos ^2}{60^ \circ }{\sin ^2}{30^ \circ } + {\tan ^2}{30^ \circ }{\cot ^2}{60^ \circ }$ by substituting the values from the trigonometric table and the relations between $\sin ,\cos ,\tan $ and $\cot $. We will use the formulas $\cot \theta = \dfrac{1}{{\tan \theta }}$ and some standard trigonometric identities.

Complete step by step answer:

The standard trigonometric table is:

	${0^ \circ }$	${30^ \circ }$	${45^ \circ }$	${60^ \circ }$	${90^ \circ }$
sin	$0$	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$
cos	$1$	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	$0$
tan	$0$	$\dfrac{1}{{\sqrt 3 }}$	$1$	$\sqrt 3 $	$\infty$

The given trigonometric expression is ${\cos ^2}{60^ \circ }{\sin ^2}{30^ \circ } + {\tan ^2}{30^ \circ }{\cot ^2}{60^ \circ }$.

From the table above we get the value of

$\cos {60^ \circ } = \dfrac{1}{2}$, $\sin {30^ \circ } = \dfrac{1}{2}$ and $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$

We can get $\cot {60^ \circ }$ by using the formula $\cot \theta = \dfrac{1}{{\tan \theta }}$
So, substituting the value of $\tan {60^ \circ } = \sqrt 3 $ in the above formula,
We get,
$\cot {60^ \circ } = \dfrac{1}{{\tan {{60}^ \circ }}}$
Substituting the value of $\tan {60^ \circ } = \sqrt 3 $,
$\cot {60^ \circ } = \dfrac{1}{{\sqrt 3 }}$

Now we have all the values for the expression,
${\cos ^2}{60^ \circ }{\sin ^2}{30^ \circ } + {\tan ^2}{30^ \circ }{\cot ^2}{60^ \circ }$

Substituting all the values from above,
$ \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} \times {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} \times {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2}$

Squaring all the terms,
$ \Rightarrow \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{1}{4}} \right) + \left( {\dfrac{1}{3}} \right) \times \left( {\dfrac{1}{3}} \right)$

Solving the multiplication terms,
$ \Rightarrow \left( {\dfrac{1}{{16}}} \right) + \left( {\dfrac{1}{9}} \right)$

Cross multiplying and taking the lowest common factor,
$ \Rightarrow \dfrac{{\left( {1 \times 9} \right) + \left( {16 \times 1} \right)}}{{16 \times 9}}$

Solving the multiplication terms,
$ \Rightarrow \dfrac{{9 + 16}}{{144}}$

Adding the numerator terms,
$ \Rightarrow \dfrac{{25}}{{144}}$

Therefore, we get ${\cos ^2}{60^ \circ }{\sin ^2}{30^ \circ } + {\tan ^2}{30^ \circ }{\cot ^2}{60^ \circ } = \dfrac{{25}}{{144}}$

Note:
We should know the difference between $sin^2\theta$ and $sin\theta^2$. $sin^2\theta$ means $(sin\theta)^2$ whereas $sin\theta^2$ means the angle is squared not the value of the sine is squared. Also we should know the signs of trigonometric functions in the four quadrants. Sine and cosecant are positive in the second quadrant and the rest are negative in the second quadrant and tangent and cotangent are positive in the third quadrant and the remaining are negative. In the fourth quadrant only cosine and secant are positive and remaining are negative.