How do you solve \[\sec x = - 5\]?
Answer
630.6k+ views
Hint: Here the question is related to the trigonometry, we use the trigonometry ratios and we are to solve this question. In this question we have to simplify the given trigonometric ratios to its simplest form. By using the trigonometry ratios and table of trigonometry ratios for standard angles we find the value for trigonometric function.
Complete step-by-step solution:
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. In trigonometry the cosecant trigonometry ratio is the reciprocal to the sine trigonometry ratio. The secant trigonometry ratio is the reciprocal to the cosine trigonometry ratio and the cotangent trigonometry ratio is the reciprocal to the tangent trigonometry ratio.
The tangent trigonometry ratio is defined as \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] , The cosecant trigonometry ratio is defined as \[\csc x = \dfrac{1}{{\sin x}}\], The secant trigonometry ratio is defined as \[\sec x = \dfrac{1}{{\cos x}}\] and The tangent trigonometry ratio is defined as \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
Now consider the given function \[\sec x = - 5\].
The secant trigonometry ratio is defined as \[\sec x = \dfrac{1}{{\cos x}}\], the given function is written as \[ \Rightarrow \dfrac{1}{{\cos x}} = - 5\]
By taking the reciprocal to the above equation we have
\[ \Rightarrow \cos x = \dfrac{{ - 1}}{5}\]
So we have
\[ \Rightarrow x = {\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{5}} \right)\]
So it is rewritten as
\[ \Rightarrow x = {180^ \circ } - {\cos ^{ - 1}}\left( {\dfrac{1}{5}} \right)\]
By the Clark’s table we have the value for \[{\cos ^{ - 1}}\left( {\dfrac{1}{5}} \right)\], so on substituting we get
\[ \Rightarrow x = {180^ \circ } - {78.46^ \circ }\]
On simplifying we get
\[ \Rightarrow x = {101.54^ \circ }\]
Note: In the trigonometry we have six trigonometry ratios and 3 trigonometry standard identities. The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, cosec or csc, sec and cot. The above question is also solved by using the relation of trigonometry ratios and the value for the trigonometry ratios in Clark's table.
Complete step-by-step solution:
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. In trigonometry the cosecant trigonometry ratio is the reciprocal to the sine trigonometry ratio. The secant trigonometry ratio is the reciprocal to the cosine trigonometry ratio and the cotangent trigonometry ratio is the reciprocal to the tangent trigonometry ratio.
The tangent trigonometry ratio is defined as \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] , The cosecant trigonometry ratio is defined as \[\csc x = \dfrac{1}{{\sin x}}\], The secant trigonometry ratio is defined as \[\sec x = \dfrac{1}{{\cos x}}\] and The tangent trigonometry ratio is defined as \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
Now consider the given function \[\sec x = - 5\].
The secant trigonometry ratio is defined as \[\sec x = \dfrac{1}{{\cos x}}\], the given function is written as \[ \Rightarrow \dfrac{1}{{\cos x}} = - 5\]
By taking the reciprocal to the above equation we have
\[ \Rightarrow \cos x = \dfrac{{ - 1}}{5}\]
So we have
\[ \Rightarrow x = {\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{5}} \right)\]
So it is rewritten as
\[ \Rightarrow x = {180^ \circ } - {\cos ^{ - 1}}\left( {\dfrac{1}{5}} \right)\]
By the Clark’s table we have the value for \[{\cos ^{ - 1}}\left( {\dfrac{1}{5}} \right)\], so on substituting we get
\[ \Rightarrow x = {180^ \circ } - {78.46^ \circ }\]
On simplifying we get
\[ \Rightarrow x = {101.54^ \circ }\]
Note: In the trigonometry we have six trigonometry ratios and 3 trigonometry standard identities. The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, cosec or csc, sec and cot. The above question is also solved by using the relation of trigonometry ratios and the value for the trigonometry ratios in Clark's table.
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