Question

Without using trigonometric tables, evaluate the following:
$\sec {41^ \circ }.\sin {49^ \circ } + \cos {49^ \circ }.\cos ec{41^ \circ } - \dfrac{2}{{\sqrt 3 }}(\tan {20^ \circ }.\tan {60^ \circ }.\tan {70^ \circ })$

Answer 1

Hint: Use the trigonometric identities-
$\cos ec({90^ \circ } - A) = \sec A,\tan A = \cot ({90^ \circ } - A)$ and then solve the question.
We have been given, $\sec {41^ \circ }.\sin {49^ \circ } + \cos {49^ \circ }.\cos ec{41^ \circ } - \dfrac{2}{{\sqrt 3 }}(\tan {20^ \circ }.\tan {60^ \circ }.\tan {70^ \circ })$.

Complete step-by-step answer:

Now using the trigonometric identities-
 $
  \sec ({41^ \circ }) = \cos ec({90^ \circ } - {41^ \circ }) \\
  \cos ec({41^ \circ }) = \sec ({90^ \circ } - {41^ \circ }) \\
  \tan ({70^ \circ }) = \cot ({90^ \circ } - 70) \\
$
So, the expression will be transformed in-
$
  \cos ec({90^ \circ } - {41^ \circ }).\sin {49^ \circ } + \cos {49^ \circ }.\sec ({90^ \circ } - {41^ \circ }) - \dfrac{2}{{\sqrt 3 }}(\tan {20^ \circ }.\tan {60^ \circ }.\cot ({90^ \circ } - {70^ \circ })) \\
   = \cos ec{49^ \circ }.\sin {49^ \circ } + \cos {49^ \circ }\sec {49^ \circ } - \dfrac{2}{{\sqrt 3 }}(\tan {20^ \circ }.\tan {60^ \circ }.\cot ({20^ \circ }) - (1) \\
$
Now, we know-
$\cos ecA = \dfrac{1}{{\sin A}},\sec A = \dfrac{1}{{\cos A}},\cot A = \dfrac{1}{{\tan A}}$, using these trigonometric formulas in equation (1), we get-
$
   = \dfrac{1}{{\sin {{49}^ \circ }}}.\sin {49^ \circ } + \cos {49^ \circ }.\dfrac{1}{{\cos {{49}^ \circ }}} - \dfrac{2}{{\sqrt 3 }}(\tan {20^ \circ }.\tan {60^ \circ }.\dfrac{1}{{\tan {{20}^ \circ }}}) \\
   = 1 + 1 - \dfrac{2}{{\sqrt 3 }}(\tan {60^ \circ }) \\
   = 1 + 1 - \dfrac{2}{{\sqrt 3 }}.\sqrt 3 \\
   = 2 - 2 \\
   = 0 \\
 $
Hence, the value of the given expression is 0.

Note: Whenever such types of question appear, then wrote down the expression given in the question and then try to convert it into a simplified form by using trigonometric formulas and then by using $\cos ecA = \dfrac{1}{{\sin A}},\sec A = \dfrac{1}{{\cos A}},\cot A = \dfrac{1}{{\tan A}}$in equation (1), we will get $ = 1 + 1 - \dfrac{2}{{\sqrt 3 }}(\tan {60^ \circ })$, solving it further we get the answer as 0.