
How do you simplify $ \dfrac{{1 - \cos {{100}^ \circ }}}{{\sin {{100}^ \circ }}} $ ?
Answer
548.1k+ views
Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $ \cos 2x = 1 - 2{\sin ^2}x $ and $ \sin 2x = 2\sin x\cos x $ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem.
Complete step-by-step answer:
In the given problem, we have to simplify the trigonometric expression $ \dfrac{{1 - \cos {{100}^ \circ }}}{{\sin {{100}^ \circ }}} $ .
So, $ \dfrac{{1 - \cos {{100}^ \circ }}}{{\sin {{100}^ \circ }}} $
Using the double angle formula of cosine $ \cos 2x = 1 - 2{\sin ^2}x $ in numerator, we get,
$ \Rightarrow $ $ \dfrac{{1 - \left( {1 - 2{{\sin }^2}{{50}^ \circ }} \right)}}{{\sin {{100}^ \circ }}} $
Using the double angle formula of sine $ \sin 2x = 2\sin x\cos x $ in denominator, we get,
$ \Rightarrow $ $ \dfrac{{1 - \left( {1 - 2{{\sin }^2}{{50}^ \circ }} \right)}}{{2\sin {{50}^ \circ }\cos {{50}^ \circ }}} $
Opening the bracket and simplifying the numerator, we get,
$ \Rightarrow $ $ \dfrac{{2{{\sin }^2}{{50}^ \circ }}}{{2\sin {{50}^ \circ }\cos {{50}^ \circ }}} $
Now, cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow $ $ \dfrac{{\sin {{50}^ \circ }}}{{\cos {{50}^ \circ }}} $
Now, we know that $ \tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}} $ . So, we get,
$ \Rightarrow \tan {50^ \circ } $
Hence, the simplification of the given trigonometric expression $ \dfrac{{1 - \cos {{100}^ \circ }}}{{\sin {{100}^ \circ }}} $ can be simplified as $ \tan {50^ \circ } $ by the use of basic algebraic rules and simple trigonometric formulae like double angle formulae for sine and cosine.
So, the correct answer is “ $ \tan {50^ \circ } $ ”.
Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: $ \tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}} $ and $ \cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}} $ . Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers. The answers may also be verified by working the solution backwards and getting the question back.
Complete step-by-step answer:
In the given problem, we have to simplify the trigonometric expression $ \dfrac{{1 - \cos {{100}^ \circ }}}{{\sin {{100}^ \circ }}} $ .
So, $ \dfrac{{1 - \cos {{100}^ \circ }}}{{\sin {{100}^ \circ }}} $
Using the double angle formula of cosine $ \cos 2x = 1 - 2{\sin ^2}x $ in numerator, we get,
$ \Rightarrow $ $ \dfrac{{1 - \left( {1 - 2{{\sin }^2}{{50}^ \circ }} \right)}}{{\sin {{100}^ \circ }}} $
Using the double angle formula of sine $ \sin 2x = 2\sin x\cos x $ in denominator, we get,
$ \Rightarrow $ $ \dfrac{{1 - \left( {1 - 2{{\sin }^2}{{50}^ \circ }} \right)}}{{2\sin {{50}^ \circ }\cos {{50}^ \circ }}} $
Opening the bracket and simplifying the numerator, we get,
$ \Rightarrow $ $ \dfrac{{2{{\sin }^2}{{50}^ \circ }}}{{2\sin {{50}^ \circ }\cos {{50}^ \circ }}} $
Now, cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow $ $ \dfrac{{\sin {{50}^ \circ }}}{{\cos {{50}^ \circ }}} $
Now, we know that $ \tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}} $ . So, we get,
$ \Rightarrow \tan {50^ \circ } $
Hence, the simplification of the given trigonometric expression $ \dfrac{{1 - \cos {{100}^ \circ }}}{{\sin {{100}^ \circ }}} $ can be simplified as $ \tan {50^ \circ } $ by the use of basic algebraic rules and simple trigonometric formulae like double angle formulae for sine and cosine.
So, the correct answer is “ $ \tan {50^ \circ } $ ”.
Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: $ \tan (x) = \dfrac{{\sin (x)}}{{\cos (x)}} $ and $ \cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}} $ . Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers. The answers may also be verified by working the solution backwards and getting the question back.
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