Answer
Verified
491.7k+ views
Hint: Use the formulas of the trigonometric functions. These trigonometric formulas are $\tan x=\cot \left( 90-x \right)$, $\cot x=\dfrac{1}{\tan x}$ and $\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1$. Here, the angle x is in degree. Using these formulas, we can solve this question.
Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In trigonometry, we have some formulas that relate tan and cot functions. These formulas are,
$\tan x=\cot \left( 90-x \right)............\left( 1 \right)$
$\cot x=\dfrac{1}{\tan x}...............\left( 2 \right)$
Also, in trigonometry we have an identity,
$\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1.................\left( 3 \right)$
In all the above 3 formulas, the angle x should be in degree.
In the question, we are required to evaluate \[\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }}\]
In formula $\left( 1 \right)$, substituting x = 53, we can write \[\tan {{53}^{\circ }}=\cot \left( 90-53 \right)=\cot {{37}^{\circ }}\] and substituting x = 77, we can write \[\tan {{77}^{\circ }}=\cot \left( 90-77 \right)=\cot {{13}^{\circ }}\]. Substituting \[\tan {{53}^{\circ }}\] and \[\tan {{77}^{\circ }}\] in the above expression, we get,
\[\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\cot {{37}^{\circ }}\cot {{13}^{\circ }}\]
In formula $\left( 2 \right)$, substituting x = 13, we get\[\cot {{13}^{\circ }}=\dfrac{1}{\tan {{13}^{\circ }}}\] and substituting x = 37, we get \[\cot {{37}^{\circ }}=\dfrac{1}{\tan {{37}^{\circ }}}\]. Substituting these in the above expression, we get,
\[\begin{align}
& \dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\dfrac{1}{\tan {{37}^{\circ }}}\dfrac{1}{\tan {{13}^{\circ }}} \\
& \Rightarrow \dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{45}^{\circ }} \\
& \Rightarrow \dfrac{2}{3}\left( \text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\cot {{58}^{\circ }}\tan {{32}^{\circ }} \right)-\dfrac{5}{3}\tan {{45}^{\circ }} \\
\end{align}\]
Using formula $\left( 1 \right)$, substituting $\tan {{32}^{\circ }}=\cot \left( 90-32 \right)=\cot {{32}^{\circ }}$ in the above expression, we get,
\[\dfrac{2}{3}\left( \text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }} \right)-\dfrac{5}{3}\tan {{45}^{\circ }}\]
From formula $\left( 3 \right)$, substituting x = 58, we get,
\[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1\]
Substituting \[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1\] in the above expression, we get,
\[\begin{align}
& \dfrac{2}{3}\left( 1 \right)-\dfrac{5}{3}\tan {{45}^{\circ }} \\
& \Rightarrow \dfrac{2}{3}-\dfrac{5}{3}\tan {{45}^{\circ }} \\
\end{align}\]
Also, from trigonometry, we have \[\tan {{45}^{\circ }}=1\]. Substituting \[\tan {{45}^{\circ }}=1\] in the above equation, we get,
\[\begin{align}
& \dfrac{2}{3}-\dfrac{5}{3}\left( 1 \right) \\
& \Rightarrow \dfrac{2}{3}-\dfrac{5}{3} \\
& \Rightarrow \dfrac{-3}{3} \\
& \Rightarrow -1 \\
\end{align}\]
Hence, the answer of the trigonometric expression is -1.
Note: There is a possibility that one may commit a mistake while using the identity \[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1\]. It is a very common mistake that one may write this as \[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}+{{\cot }^{2}}{{58}^{\circ }}=1\] instead of \[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1\]. To avoid this mistake, one should start from the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and divide both sides of the equation by ${{\sin }^{2}}x$. Then substitute $\dfrac{\cos x}{\sin x}=1$ and $\dfrac{1}{\sin x}=\text{cosecx}$. Substitute $\dfrac{\cos x}{\sin x}=1$ and $\dfrac{1}{\sin x}=\text{cosecx}$ in ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we will get the identity $\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1$.
Complete step-by-step answer:
Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In trigonometry, we have some formulas that relate tan and cot functions. These formulas are,
$\tan x=\cot \left( 90-x \right)............\left( 1 \right)$
$\cot x=\dfrac{1}{\tan x}...............\left( 2 \right)$
Also, in trigonometry we have an identity,
$\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1.................\left( 3 \right)$
In all the above 3 formulas, the angle x should be in degree.
In the question, we are required to evaluate \[\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\tan {{53}^{\circ }}\tan {{77}^{\circ }}\]
In formula $\left( 1 \right)$, substituting x = 53, we can write \[\tan {{53}^{\circ }}=\cot \left( 90-53 \right)=\cot {{37}^{\circ }}\] and substituting x = 77, we can write \[\tan {{77}^{\circ }}=\cot \left( 90-77 \right)=\cot {{13}^{\circ }}\]. Substituting \[\tan {{53}^{\circ }}\] and \[\tan {{77}^{\circ }}\] in the above expression, we get,
\[\dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\cot {{37}^{\circ }}\cot {{13}^{\circ }}\]
In formula $\left( 2 \right)$, substituting x = 13, we get\[\cot {{13}^{\circ }}=\dfrac{1}{\tan {{13}^{\circ }}}\] and substituting x = 37, we get \[\cot {{37}^{\circ }}=\dfrac{1}{\tan {{37}^{\circ }}}\]. Substituting these in the above expression, we get,
\[\begin{align}
& \dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{13}^{\circ }}\tan {{37}^{\circ }}\tan {{45}^{\circ }}\dfrac{1}{\tan {{37}^{\circ }}}\dfrac{1}{\tan {{13}^{\circ }}} \\
& \Rightarrow \dfrac{2}{3}\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\dfrac{2}{3}\cot {{58}^{\circ }}\tan {{32}^{\circ }}-\dfrac{5}{3}\tan {{45}^{\circ }} \\
& \Rightarrow \dfrac{2}{3}\left( \text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-\cot {{58}^{\circ }}\tan {{32}^{\circ }} \right)-\dfrac{5}{3}\tan {{45}^{\circ }} \\
\end{align}\]
Using formula $\left( 1 \right)$, substituting $\tan {{32}^{\circ }}=\cot \left( 90-32 \right)=\cot {{32}^{\circ }}$ in the above expression, we get,
\[\dfrac{2}{3}\left( \text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }} \right)-\dfrac{5}{3}\tan {{45}^{\circ }}\]
From formula $\left( 3 \right)$, substituting x = 58, we get,
\[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1\]
Substituting \[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1\] in the above expression, we get,
\[\begin{align}
& \dfrac{2}{3}\left( 1 \right)-\dfrac{5}{3}\tan {{45}^{\circ }} \\
& \Rightarrow \dfrac{2}{3}-\dfrac{5}{3}\tan {{45}^{\circ }} \\
\end{align}\]
Also, from trigonometry, we have \[\tan {{45}^{\circ }}=1\]. Substituting \[\tan {{45}^{\circ }}=1\] in the above equation, we get,
\[\begin{align}
& \dfrac{2}{3}-\dfrac{5}{3}\left( 1 \right) \\
& \Rightarrow \dfrac{2}{3}-\dfrac{5}{3} \\
& \Rightarrow \dfrac{-3}{3} \\
& \Rightarrow -1 \\
\end{align}\]
Hence, the answer of the trigonometric expression is -1.
Note: There is a possibility that one may commit a mistake while using the identity \[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1\]. It is a very common mistake that one may write this as \[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}+{{\cot }^{2}}{{58}^{\circ }}=1\] instead of \[\text{cose}{{\text{c}}^{2}}{{58}^{\circ }}-{{\cot }^{2}}{{58}^{\circ }}=1\]. To avoid this mistake, one should start from the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and divide both sides of the equation by ${{\sin }^{2}}x$. Then substitute $\dfrac{\cos x}{\sin x}=1$ and $\dfrac{1}{\sin x}=\text{cosecx}$. Substitute $\dfrac{\cos x}{\sin x}=1$ and $\dfrac{1}{\sin x}=\text{cosecx}$ in ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we will get the identity $\text{cose}{{\text{c}}^{2}}x-{{\cot }^{2}}x=1$.
Recently Updated Pages
Identify the feminine gender noun from the given sentence class 10 english CBSE
Your club organized a blood donation camp in your city class 10 english CBSE
Choose the correct meaning of the idiomphrase from class 10 english CBSE
Identify the neuter gender noun from the given sentence class 10 english CBSE
Choose the word which best expresses the meaning of class 10 english CBSE
Choose the word which is closest to the opposite in class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
A rainbow has circular shape because A The earth is class 11 physics CBSE
The male gender of Mare is Horse class 11 biology CBSE
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths