Question

Find the value of ${\tan ^2}{60^ \circ }\cos e{c^2}{45^ \circ } + {\sec ^2}{45^ \circ }\sin {30^ \circ }$
\[{\text{A}}{\text{.}}\]-7
${\text{B}}{\text{.}}$ 6
${\text{C}}{\text{.}}$ -6
${\text{D}}{\text{.}}$ 7

Answer 1

Hint- Use the values, $\tan {60^ \circ } = \sqrt 3 $; $\cos ec{45^ \circ } = \sqrt 2 $; $\sec {45^ \circ } = \sqrt 2 $ and $\sin {30^ \circ } = \dfrac{1}{2}$. Put these values in the given equation in question and find the answer.

Complete step-by-step answer:
Let us assume $S = {\tan ^2}{60^ \circ }\cos e{c^2}{45^ \circ } + {\sec ^2}{45^ \circ }\sin {30^ \circ }$.
Now as we know the values $\tan {60^ \circ } = \sqrt 3 $; $\cos ec{45^ \circ } = \sqrt 2 $; $\sec {45^ \circ } = \sqrt 2 $ and $\sin {30^ \circ } = \dfrac{1}{2}$.
We can directly substitute these values in the given equation and find the answer.
Put the values, we get-
$S = {\tan ^2}{60^ \circ }\cos e{c^2}{45^ \circ } + {\sec ^2}{45^ \circ }\sin {30^ \circ }$
$ \Rightarrow {\left( {\sqrt 3 } \right)^2}.{\left( {\sqrt 2 } \right)^2} + {\left( {\sqrt 2 } \right)^2}.\dfrac{1}{2}$
Solving further we will get-
$
   = 3 \times 2 + 2 \times \dfrac{1}{2} \\
   = 6 + 1 \\
   = 7 \\
 $
We get,
$S = {\tan ^2}{60^ \circ }\cos e{c^2}{45^ \circ } + {\sec ^2}{45^ \circ }\sin {30^ \circ } = 7$
Hence, the answer to the given question is 7, so, the correct option is option (D).
This is the simplest way to solve the equation.

Note – In such types of questions, first write the equation given in the question whose value is to be found out. Assume the equation given in the question to be S. Put the values of the terms in the equation by using the trigonometric table. So, after putting the values of the $\tan {60^ \circ }$ ; $\cos ec{45^ \circ }$ ; \[\sec {45^ \circ }\] and $\sin {30^ \circ }$ and then evaluating the equation we get the answer as 7. But you have to be careful while keeping the values, don’t write incorrect values as it can lead to incorrect answers to the given solution.