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Question: Show that \(\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \...

Show that tan⁡57∘+cot⁡37∘tan⁡33∘+cot⁡53∘\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}} is equal to
A. tan⁡33∘cot⁡57∘\tan {33^ \circ }\cot {57^ \circ }
B. tan⁡57∘cot⁡37∘\tan {57^ \circ }\cot {37^ \circ }
C. tan⁡33∘cot⁡53∘\tan {33^ \circ }\cot {53^ \circ }
D. tan⁡53∘cot⁡37∘\tan {53^ \circ }\cot {37^ \circ }

Explanation

Solution

As we can see that the above question is related to trigonometry as tangent and cotangent are trigonometric ratios. WE can solve this question by applying the trigonometric identities. Some of the basic identities are tan⁡(90−θ)=cot⁡θ\tan (90 - \theta ) = \cot \theta and we can write cot⁡(90−ϕ)=tan⁡ϕ\cot (90 - \phi ) = \tan \phi . We should also know that cot⁡θ\cot \theta can be written as 1tan⁡θ\dfrac{1}{{\tan \theta }}.

Complete step by step answer:
Here we have tan⁡57∘+cot⁡37∘tan⁡33∘+cot⁡53∘\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{33}^ \circ } + \cot {{53}^ \circ }}}.
By applying the identities tan⁡(90−θ)=cot⁡θ\tan (90 - \theta ) = \cot \theta and cot⁡(90−ϕ)=tan⁡ϕ\cot (90 - \phi ) = \tan \phi in the denominator we can solve this. By comparing for tangent we have θ=57\theta = 57, and for cotangent we have ϕ=37\phi = 37.
So we can write tan⁡57∘+cot⁡37∘tan⁡(90−57)∘+cot⁡(90−37)∘\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\tan {{(90 - 57)}^ \circ } + \cot {{(90 - 37)}^ \circ }}}. It can be written as tan⁡57∘+cot⁡37∘cot⁡57∘+tan⁡37∘\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\cot {{57}^ \circ } + \tan {{37}^ \circ }}}.
We know that cot⁡ϕ\cot \phi can be written as 1tan⁡ϕ\dfrac{1}{{\tan \phi }} and the same for tan θ\theta .
We can write the expression as tan⁡57∘+cot⁡37∘1tan⁡57∘+1cot⁡37∘\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\dfrac{1}{{\tan {{57}^ \circ }}} + \dfrac{1}{{\cot {{37}^ \circ }}}}}.
By taking the LCM of the denominator of the denominator we can write tan⁡57∘+cot⁡37∘cot⁡37+tan⁡57tan⁡57∘⋅cot⁡37∘\dfrac{{\tan {{57}^ \circ } + \cot {{37}^ \circ }}}{{\dfrac{{\cot 37 + \tan 57}}{{\tan {{57}^ \circ } \cdot \cot {{37}^ \circ }}}}}.
On further solving we can write tan⁡57∘⋅cot⁡37[tan⁡57+cot⁡37tan⁡57+cot⁡37]\tan {57^ \circ } \cdot \cot 37\left[ {\dfrac{{\tan 57 + \cot 37}}{{\tan 57 + \cot 37}}} \right].
So it gives us the value tan⁡57∘⋅cot⁡37∘\tan {57^ \circ } \cdot \cot {37^ \circ }.

Hence, the correct answer is option B.

Note: We should note that tan⁡θ\tan \theta can also be written as 1cot⁡θ\dfrac{1}{{\cot \theta }}. Before solving this kind of question we should have the full knowledge of trigonometric functions and their identities. We know that if there is abc\dfrac{a}{{\dfrac{b}{c}}}, then it can be written as c×ab\dfrac{{c \times a}}{b}. This is what we have applied in the above solution.