Question

Verify the following trigonometric expression
\[\dfrac{{cotx - tanx}}{{\sin x\cos x}} = cose{c^2}x - {\sec ^2}x\]

Answer 1

Hint: In order to verify the expression we will take the complex side of the equation, simplify it using reciprocal and quotient identities. The nature of the identities will depend on the nature of the question. After successful application the identities we will simplify the question until we get the same expression as on the other side.

Complete step-by-step solution:
We have,
\[\dfrac{{cotx - tanx}}{{\sin x\cos x}} = cose{c^2}x - {\sec ^2}x\]
Choosing the complex side, that is, LHS
We have,
\[
  \dfrac{{cotx - tanx}}{{\sin x\cos x}} - - - - - \left( 1 \right) \\
    \\
    \\
 \]
According to the quotient identities of trigonometry, we know that,
\[\tan x = \dfrac{{\sin x}}{{\cos x}} - - - - \left( a \right)\]
And
\[cotx = \dfrac{{\cos x}}{{\sin x}} - - - - - \left( b \right)\]
Substituting the values of\[\left( a \right)\]and \[\left( b \right)\]in \[\left( 1 \right)\]
We get,
\[ = \dfrac{{\dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}}}}{{\sin x\cos x}}\]
Taking L.C.M. of numerator, we get
\[ = \dfrac{{\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x\cos x}}}}{{\sin x\cos x}}\]
\[ \Rightarrow \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\sin x\cos x} \right)}^2}}}\]
Further we split the equation,
\[ \Rightarrow \dfrac{{{{\cos }^2}x}}{{{{\left( {\sin x\cos x} \right)}^2}}} - \dfrac{{si{n^2}x}}{{{{\left( {\sin x\cos x} \right)}^2}}}\]
Therefore,
\[ \Rightarrow \dfrac{{{{\cos }^2}x}}{{\left( {{{\sin }^2}x{{\cos }^{}}x} \right)}} - \dfrac{{si{n^2}x}}{{\left( {{{\sin }^2}x{{\cos }^2}x} \right)}}\]
After cutting the common factors, we get,
\[ \Rightarrow \dfrac{1}{{{{\sin }^2}x}} - \dfrac{1}{{{{\cos }^2}x}}\]
And according to the reciprocal identities of trigonometric function, we know
\[\dfrac{1}{{{{\sin }^2}x}} = \cos e{c^2}x\] And \[\dfrac{1}{{{{\cos }^2}x}} = {\sec ^2}x\]

Thus, \[\dfrac{1}{{{{\sin }^2}x}} - \dfrac{1}{{{{\cos }^2}x}} = \cos e{c^2}x - {\sec ^2}x = RHS\]
Hence, Verified.

Additional information: the alternative way of solving the same problem would be starting with Right hand side of the equation and substituting the values from the following trigonometric identities, \[1 + {\tan ^2}x = se{c^2}x\] and \[1 + {\cot ^2}x = cose{c^2}x\]. Further simplification would involve use of quotient identities. The quotient identities determine the relationship between various trigonometric functions. They are as follows:
\[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
\[\cot x = \dfrac{{\cos x}}{{\sin x}}\]

Note: To prove trigonometric equations we should usually start with the complicated side of the question and keep simplifying the equation until it has been transformed into the same expression as on the other side. The other methods of solving a trigonometric equation involve expanding the expressions, factoring the expression or simply using basic algebraic strategies to obtain desired results. Simplifying one side of the equation equal to the other side is one of the most commonly used methods.