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Question: Without using trigonometric tables, evaluate the following: \(\sec {41^ \circ }.\sin {49^ \circ } ...

Without using trigonometric tables, evaluate the following:
sec⁡41∘.sin⁡49∘+cos⁡49∘.cos⁡ec41∘−23(tan⁡20∘.tan⁡60∘.tan⁡70∘)\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 })

Explanation

Solution

Hint: Use the trigonometric identities-
cos⁡ec(90∘−A)=sec⁡A,tan⁡A=cot⁡(90∘−A)\cos ec({90^ \circ } - A) = \sec A,\tan A = \cot ({90^ \circ } - A) and then solve the question.
We have been given, sec⁡41∘.sin⁡49∘+cos⁡49∘.cos⁡ec41∘−23(tan⁡20∘.tan⁡60∘.tan⁡70∘)\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∘)=cos⁡ec(90∘−41∘) cos⁡ec(41∘)=sec⁡(90∘−41∘) tan⁡(70∘)=cot⁡(90∘−70)  \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∘−41∘).sin⁡49∘+cos⁡49∘.sec⁡(90∘−41∘)−23(tan⁡20∘.tan⁡60∘.cot⁡(90∘−70∘)) =cos⁡ec49∘.sin⁡49∘+cos⁡49∘sec⁡49∘−23(tan⁡20∘.tan⁡60∘.cot⁡(20∘)−(1)  \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=1sin⁡A,sec⁡A=1cos⁡A,cot⁡A=1tan⁡A\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-
=1sin⁡49∘.sin⁡49∘+cos⁡49∘.1cos⁡49∘−23(tan⁡20∘.tan⁡60∘.1tan⁡20∘) =1+1−23(tan⁡60∘) =1+1−23.3 =2−2 =0  = \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=1sin⁡A,sec⁡A=1cos⁡A,cot⁡A=1tan⁡A\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+1−23(tan⁡60∘) = 1 + 1 - \dfrac{2}{{\sqrt 3 }}(\tan {60^ \circ }), solving it further we get the answer as 0.