Question

How do you prove \[\tan x + \cot x = \sec x\csc x\]?

Answer 1

Hint: Here, we will take into consideration the left-hand side of the given equation. We will then convert the trigonometric function in terms of sine and cosine functions. We will then take LCM of the terms and then simplify using trigonometric ratios and reciprocal identities to prove the equation.

Formula Used:
We will use the following formulas:
1. \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
2. \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]

Complete step-by-step solution:
We will use trigonometric ratios and identities to simplify the left-hand side.
The left hand side of the given equation is \[\tan x + \cot x\].
The tangent of an angle \[x\] can be written as the quotient of the sine and cosine of the angle \[x\]. This can be written as \[\tan x = \dfrac{{\sin x}}{{\cos x}}\].
The cotangent of an angle \[x\] can be written as the quotient of the cosine and sine of the angle \[x\]. This can be written as \[\cot x = \dfrac{{\cos x}}{{\sin x}}\].
Substituting \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] and \[\cot x = \dfrac{{\cos x}}{{\sin x}}\] in the left hand side of the equation, we get
\[\begin{array}{l}{\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \tan x + \cot x\\ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}}\end{array}\]
The lowest common denominator of the denominators \[\sin x\] and \[\cos x\] is \[\sin x\cos x\].
Multiplying and dividing \[\dfrac{{\sin x}}{{\cos x}}\] by \[\sin x\], and \[\dfrac{{\cos x}}{{\sin x}}\] by \[\cos x\], we get
\[\begin{array}{c} \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{\sin x}}{{\cos x}} \cdot \dfrac{{\sin x}}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}} \cdot \dfrac{{\cos x}}{{\cos x}}\\ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{{{\sin }^2}x}}{{\sin x\cos x}} + \dfrac{{{{\cos }^2}x}}{{\sin x\cos x}}\end{array}\]
Taking the lowest common denominator, we get
\[ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x\cos x}}\]
The sum of the squares of the sine and cosine of an angle is always equal to 1.
Therefore, we get
\[{\sin ^2}x + {\cos ^2}x = 1\]
Substituting \[{\sin ^2}x + {\cos ^2}x = 1\] in the L.H.S., we get
\[ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{\sin x\cos x}}\]
Rewriting the equation, we get
\[ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{\sin x}} \times \dfrac{1}{{\cos x}}\]
The secant of an angle \[x\] can be written as the reciprocal of cosine of the angle \[x\]. This can be written as \[\sec x = \dfrac{1}{{\cos x}}\].
The cosecant of an angle \[x\] can be written as the reciprocal of sine of the angle \[x\]. This can be written as \[\csc x = \dfrac{1}{{\sin x}}\].
Substituting \[\dfrac{1}{{\cos x}} = \sec x\] and \[\dfrac{1}{{\sin x}} = \csc x\] in the L.H.S., we get
\[\begin{array}{c} \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \csc x\sec x\\ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \sec x\csc x\end{array}\]
Therefore, we get
\[{\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}\]

Hence, we have proved the equation \[\tan x + \cot x = \sec x\csc x\].

Note:
Alternate method: We can simplify the right-hand side to prove the equation.
Rewriting the right hand side, we get
\[\begin{array}{l}{\text{R}}{\text{.H}}{\text{.S}}{\text{.}} = \sec x\csc x\\ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{\cos x}} \times \dfrac{1}{{\sin x}}\\ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{\sin x\cos x}}\end{array}\]
Substituting \[1 = {\sin ^2}x + {\cos ^2}x\], we get
\[ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x\cos x}}\]
Splitting the L.C.M., we get
\[ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{{{\sin }^2}x}}{{\sin x\cos x}} + \dfrac{{{{\cos }^2}x}}{{\sin x\cos x}}\]
Simplifying the expression by removing the common factor from numerator and denominator, we get
\[ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}}\]
Therefore, we get
\[ \Rightarrow {\text{R}}{\text{.H}}{\text{.S}}{\text{.}} = \tan x + \cot x = L.H.S.\]
Hence, we have proved the equation \[\tan x + \cot x = \sec x\csc x\].