Question

If a trigonometric expression is given as \[{\text{cosec}}x + \cot x = p\], then prove that \[\cos x = \dfrac{{{p^2} - 1}}{{{p^2} + 1}}\].

Answer 1

Hint: Recall the trigonometric relations between the functions sine and cosine which is \[{\sin ^2}x + {\cos ^2}x = 1\] and use them to simplify the expression.

Complete step-by-step answer:
Trigonometric functions are defined on the angles that relate the ratio of the sides of the triangle. Six different combinations of three different sides are possible and hence, we have six trigonometric functions.
The sine, cosine, and tangent are primary and cosecant, secant, and cotangent are inverses of them respectively.
It is given that \[{\text{cosec}}x + \cot x = p\] and we need to prove that \[\cos x = \dfrac{{{p^2} - 1}}{{{p^2} + 1}}\].
We can express cosecant and cotangent in terms of cosine and sine as follows:
\[{\text{cosec}}x = \dfrac{1}{{\sin x}}\]
\[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
Then, the expression for p is given as follows:
\[\dfrac{1}{{\sin x}} + \dfrac{{\cos x}}{{\sin x}} = p\]
Simplifying, we have:
\[p = \dfrac{{1 + \cos x}}{{\sin x}}...........(1)\]
Now, we try to evaluate \[\dfrac{{{p^2} - 1}}{{{p^2} + 1}}\] using equation (1) as follows:
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \dfrac{{{{\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right)}^2} - 1}}{{{{\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right)}^2} + 1}}\]
We now evaluate the square terms.
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \dfrac{{\left( {\dfrac{{1 + 2\cos x + {{\cos }^2}x}}{{{{\sin }^2}x}}} \right) - 1}}{{\left( {\dfrac{{1 + 2\cos x + {{\cos }^2}x}}{{{{\sin }^2}x}}} \right) + 1}}\]
Taking common denominator, we have:
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \dfrac{{\dfrac{{1 + 2\cos x + {{\cos }^2}x - {{\sin }^2}x}}{{{{\sin }^2}x}}}}{{\dfrac{{1 + 2\cos x + {{\cos }^2}x + {{\sin }^2}x}}{{{{\sin }^2}x}}}}\]
Canceling the denominator, we have:
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \dfrac{{1 + 2\cos x + {{\cos }^2}x - {{\sin }^2}x}}{{1 + 2\cos x + {{\cos }^2}x + {{\sin }^2}x}}\]
We know that \[{\sin ^2}x + {\cos ^2}x = 1\], using this formula in the above equation, we get:
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \dfrac{{1 + 2\cos x + {{\cos }^2}x - {{\sin }^2}x}}{{1 + 2\cos x + 1}}\]
Now, we also know that \[1 - {\sin ^2}x = {\cos ^2}x\], using this formula in the above equation, we get:
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \dfrac{{2\cos x + {{\cos }^2}x + {{\cos }^2}x}}{{2 + 2\cos x}}\]
Simplifying the numerator, we get:
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \dfrac{{2\cos x + 2{{\cos }^2}x}}{{2 + 2\cos x}}\]
Taking cosx common from the numerator of the above expression, we get:
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \dfrac{{\cos x(2 + 2\cos x)}}{{2 + 2\cos x}}\]
Canceling the common term in numerator and denominator, we get as follows:
\[\dfrac{{{p^2} - 1}}{{{p^2} + 1}} = \cos x\]
The above expression is the required proof.
Hence, we proved.

Note: You can also directly work with cosecant and cotangent functions but it is easier to evaluate once we convert it to cosine and since. Also, note that the final answer is in cosine, hence, it is convenient to convert the functions first and then evaluate.