
How do you simplify \[\dfrac{1-\cos \left( {{80}^{\circ }} \right)}{1+\cos \left( {{80}^{\circ }} \right)}\]?
Answer
559.5k+ views
Hint: In this problem, we have to simplify the given trigonometric expression and reduce it. To solve these types of problems, we should know basic trigonometric formulas and identities. In this problem we can use trigonometric identities and convert the cosines into sines and cosines to get the exact simplified form. In this problem we can use the identities like \[1+\cos 2x=2{{\cos }^{2}}x\] and \[1-\cos 2x=2{{\sin }^{2}}x\] to convert the given problem and can use the formula \[\dfrac{\sin x}{\cos x}=\tan x\] to simplify the given problem.
Complete step by step answer:
We know that the given trigonometric expression to be simplified is,
\[\dfrac{1-\cos \left( {{80}^{\circ }} \right)}{1+\cos \left( {{80}^{\circ }} \right)}\] …… (1)
We also know that the trigonometric identities, we are going to use are
\[\begin{align}
& 1-\cos 2x=2{{\sin }^{2}}x \\
& 1+\cos 2x=2{{\cos }^{2}}x \\
\end{align}\]
Now we can substitute the above formula in the expression (1), we get
\[\Rightarrow \dfrac{2{{\sin }^{2}}{{40}^{\circ }}}{2{{\cos }^{2}}{{40}^{\circ }}}\]
We know that in the expression (1), we have degree value as 2a and we can convert it to a.
\[\begin{align}
& \Rightarrow 2x={{80}^{\circ }} \\
& \Rightarrow x={{40}^{\circ }} \\
\end{align}\]
We also know that, \[\dfrac{\sin x}{\cos x}=\tan x\], applying this in the above step and cancelling similar numbers, we get,
\[\begin{align}
& \Rightarrow \dfrac{{{\sin }^{2}}{{40}^{\circ }}}{{{\cos }^{2}}{{60}^{\circ }}} \\
& \Rightarrow {{\tan }^{2}}{{40}^{\circ }} \\
\end{align}\]
Therefore, the simplified form of \[\dfrac{1-\cos \left( {{80}^{\circ }} \right)}{1+\cos \left( {{80}^{\circ }} \right)}\] is \[{{\tan }^{2}}{{40}^{\circ }}\].
Note:
Students make mistakes in writing the trigonometric identities and formulas, we should always remember the trigonometric rules, formulas, properties and identities to solve these types of problems. In this problem we have used the main trigonometric identities that is, \[1-\cos 2x=2{{\sin }^{2}}x\], \[1+\cos 2x=2{{\cos }^{2}}x\] and \[\dfrac{\sin x}{\cos x}=\tan x\] which are the basic trigonometric identities used in many problems to solve. We should also concentrate on the part where 2a is given, which must be converted into a.
Complete step by step answer:
We know that the given trigonometric expression to be simplified is,
\[\dfrac{1-\cos \left( {{80}^{\circ }} \right)}{1+\cos \left( {{80}^{\circ }} \right)}\] …… (1)
We also know that the trigonometric identities, we are going to use are
\[\begin{align}
& 1-\cos 2x=2{{\sin }^{2}}x \\
& 1+\cos 2x=2{{\cos }^{2}}x \\
\end{align}\]
Now we can substitute the above formula in the expression (1), we get
\[\Rightarrow \dfrac{2{{\sin }^{2}}{{40}^{\circ }}}{2{{\cos }^{2}}{{40}^{\circ }}}\]
We know that in the expression (1), we have degree value as 2a and we can convert it to a.
\[\begin{align}
& \Rightarrow 2x={{80}^{\circ }} \\
& \Rightarrow x={{40}^{\circ }} \\
\end{align}\]
We also know that, \[\dfrac{\sin x}{\cos x}=\tan x\], applying this in the above step and cancelling similar numbers, we get,
\[\begin{align}
& \Rightarrow \dfrac{{{\sin }^{2}}{{40}^{\circ }}}{{{\cos }^{2}}{{60}^{\circ }}} \\
& \Rightarrow {{\tan }^{2}}{{40}^{\circ }} \\
\end{align}\]
Therefore, the simplified form of \[\dfrac{1-\cos \left( {{80}^{\circ }} \right)}{1+\cos \left( {{80}^{\circ }} \right)}\] is \[{{\tan }^{2}}{{40}^{\circ }}\].
Note:
Students make mistakes in writing the trigonometric identities and formulas, we should always remember the trigonometric rules, formulas, properties and identities to solve these types of problems. In this problem we have used the main trigonometric identities that is, \[1-\cos 2x=2{{\sin }^{2}}x\], \[1+\cos 2x=2{{\cos }^{2}}x\] and \[\dfrac{\sin x}{\cos x}=\tan x\] which are the basic trigonometric identities used in many problems to solve. We should also concentrate on the part where 2a is given, which must be converted into a.
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