
Given \[\tan t=-4\] and \[\text{cosec t 0}\], how do you find the other \[5\] trig function?
Answer
557.1k+ views
Hint: These types of problems are pretty straight forward and very easy to solve. The above mentioned problem is a simple demonstration of trigonometric equations, relations and expressions. Before solving this type of problems which involve simplifying, we need to recall some of the basic equations and relations of trigonometry. We start off the solution by converting the given problem into simpler forms which have \[\sin x\] and \[\cos x\] as the main terms. The equations are described as:
\[\begin{align}
& \sec \theta =\dfrac{1}{\cos \theta } \\
& \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
& \cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
We also need to remember some of the basic defined relations, which are:
\[\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
& {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
& \text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1 \\
\end{align}\]
Complete step by step answer:
Now starting off with the solution, we can firstly notice the form given is \[\tan t\] , and on rewriting this value of \[\tan t\] as \[\tan t=\dfrac{\sin t}{\cos t}\] . In this problem the value of \[\tan t\] is given to be as negative, and the value of \[\text{cosect }\]is given to be as negative. This means that the value of \[\sin t\] must also be negative, as we know, \[\text{cosect}=\dfrac{1}{\sin t}\] . From this information we can say hence the value of \[\cos t\] and \[\sec t\] must be positive. We can conclude from this, that the angle \[t\] lies in the 4th quadrant, where the value of \[\cos t\] and \[\sec t\] is positive.
At first we find the value of the angle \[t\] . It can be found as \[t={{\tan }^{-1}}\left( -4 \right)\] . Thus, it turns out to be, \[t=-{{\tan }^{-1}}\left( 4 \right)\] . \[\Rightarrow t=-{{75.9638}^{\circ }}\] . Writing it in positive form by adding \[{{360}^{\circ }}\] to it, we get,
\[\begin{align}
& \Rightarrow t=-{{75.9638}^{\circ }}+360 \\
& \Rightarrow t={{284.0362}^{\circ }} \\
\end{align}\]
Now, finding all the other trigonometric functions, we get,
\[\begin{align}
& \sin t=\sin \left( {{284.0362}^{\circ }} \right) \\
& \Rightarrow \sin t=-0.9701 \\
\end{align}\]
We get \[\text{cosect}=\dfrac{1}{\sin t}\] , Thus,
\[\begin{align}
& \text{cosect}=\dfrac{1}{\sin t} \\
& \Rightarrow \text{cosect}=-\dfrac{1}{0.9701} \\
& \Rightarrow \text{cosect}=-1.0308 \\
\end{align}\]
Now, we know, \[\tan t=\dfrac{\sin t}{\cos t}\] . Thus from this, we can write \[\cos t\] as,
\[\begin{align}
& \cos t=\dfrac{\sin t}{\tan t} \\
& \Rightarrow \cos t=\dfrac{-0.9701}{-4} \\
& \Rightarrow \cos t=\dfrac{0.9701}{4} \\
& \Rightarrow \cos t=0.2425 \\
\end{align}\]
From the relation \[\sec t=\dfrac{1}{\cos t}\] we can easily find the value of \[\sec t\] as,
\[\begin{align}
& \sec t=\dfrac{1}{\cos t} \\
& \Rightarrow \sec t=\dfrac{1}{0.2425} \\
& \Rightarrow \sec t=4.1237 \\
\end{align}\]
Note:
We should always remember all the equations and general values of trigonometry or else solving such problems may be difficult. We should also properly evaluate the value of the angle, and should be able to determine the angle in the proper quadrant with the proper sign. Once determined the quadrant, we can easily figure out the value and sign of the rest of the trigonometric values.
\[\begin{align}
& \sec \theta =\dfrac{1}{\cos \theta } \\
& \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
& \cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
We also need to remember some of the basic defined relations, which are:
\[\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
& {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
& \text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1 \\
\end{align}\]
Complete step by step answer:
Now starting off with the solution, we can firstly notice the form given is \[\tan t\] , and on rewriting this value of \[\tan t\] as \[\tan t=\dfrac{\sin t}{\cos t}\] . In this problem the value of \[\tan t\] is given to be as negative, and the value of \[\text{cosect }\]is given to be as negative. This means that the value of \[\sin t\] must also be negative, as we know, \[\text{cosect}=\dfrac{1}{\sin t}\] . From this information we can say hence the value of \[\cos t\] and \[\sec t\] must be positive. We can conclude from this, that the angle \[t\] lies in the 4th quadrant, where the value of \[\cos t\] and \[\sec t\] is positive.
At first we find the value of the angle \[t\] . It can be found as \[t={{\tan }^{-1}}\left( -4 \right)\] . Thus, it turns out to be, \[t=-{{\tan }^{-1}}\left( 4 \right)\] . \[\Rightarrow t=-{{75.9638}^{\circ }}\] . Writing it in positive form by adding \[{{360}^{\circ }}\] to it, we get,
\[\begin{align}
& \Rightarrow t=-{{75.9638}^{\circ }}+360 \\
& \Rightarrow t={{284.0362}^{\circ }} \\
\end{align}\]
Now, finding all the other trigonometric functions, we get,
\[\begin{align}
& \sin t=\sin \left( {{284.0362}^{\circ }} \right) \\
& \Rightarrow \sin t=-0.9701 \\
\end{align}\]
We get \[\text{cosect}=\dfrac{1}{\sin t}\] , Thus,
\[\begin{align}
& \text{cosect}=\dfrac{1}{\sin t} \\
& \Rightarrow \text{cosect}=-\dfrac{1}{0.9701} \\
& \Rightarrow \text{cosect}=-1.0308 \\
\end{align}\]
Now, we know, \[\tan t=\dfrac{\sin t}{\cos t}\] . Thus from this, we can write \[\cos t\] as,
\[\begin{align}
& \cos t=\dfrac{\sin t}{\tan t} \\
& \Rightarrow \cos t=\dfrac{-0.9701}{-4} \\
& \Rightarrow \cos t=\dfrac{0.9701}{4} \\
& \Rightarrow \cos t=0.2425 \\
\end{align}\]
From the relation \[\sec t=\dfrac{1}{\cos t}\] we can easily find the value of \[\sec t\] as,
\[\begin{align}
& \sec t=\dfrac{1}{\cos t} \\
& \Rightarrow \sec t=\dfrac{1}{0.2425} \\
& \Rightarrow \sec t=4.1237 \\
\end{align}\]
Note:
We should always remember all the equations and general values of trigonometry or else solving such problems may be difficult. We should also properly evaluate the value of the angle, and should be able to determine the angle in the proper quadrant with the proper sign. Once determined the quadrant, we can easily figure out the value and sign of the rest of the trigonometric values.
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