Question

How do you prove \[\left( {\sin x + \cos x} \right)(\tan x + \cot x) = \sec x + \operatorname{cs} cx?\]

Answer 1

Hint: We take left-hand side terms to prove that the right-hand side. First multiple the left-hand side terms and use the trigonometric quotient rule. We need to convert tan and cot in terms of sine and cos by using the quotient rule. Then take the least common multiple to the denominator and now take common factors to the numerator. Then we need to use the trigonometry identity formula, and the trigonometric reciprocal rule to find right-hand side terms.

Formula:
Trigonometry formula,
\[\tan x = \dfrac{{\sin x}}{{\cos x}}\] and \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
Trigonometry identity formula,
\[{\sin ^2}x + {\cos ^2}x = 1\]
Trigonometry formula,
\[\dfrac{1}{{\cos x}} = \sec x\] and \[\dfrac{1}{{\sin x}} = \operatorname{cs} cx\]

Complete answer:
First, we work to the left-hand side then find the right-hand side
The left-hand side \[ = \left( {\sin x + \cos x} \right)(\tan x + \cot x)\]
Multiple on both terms on the left-hand side
\[ = \sin x\tan x + \sin x\cos x + \cos x\tan x + \cos x\cot x\]
Now we change the \[\tan x = \dfrac{{\sin x}}{{\cos x}}\] and \[\cot x = \dfrac{{\cos x}}{{\sin x}}\]
\[ = \sin x\dfrac{{\sin x}}{{\cos x}} + \sin x\dfrac{{\cos x}}{{\sin x}} + \cos x\dfrac{{\sin x}}{{\cos x}} + \cos x\dfrac{{\cos x}}{{\sin x}}\]
Sum the denominators are same terms
\[ = \dfrac{{\left( {\sin x\sin x} \right)}}{{\cos x}} + \dfrac{{\left( {\cos x\sin x} \right)}}{{\cos x}} + \dfrac{{\left( {\sin x\cos x} \right)}}{{\sin x}} + \dfrac{{\left( {\cos x\cos x} \right)}}{{\sin x}}\]
\[ = \dfrac{{\left( {\sin x\sin x} \right) + \left( {\cos x\sin x} \right)}}{{\cos x}} + \dfrac{{\left( {\sin x\cos x} \right) + \left( {\cos x\cos x} \right)}}{{\sin x}}\]
Take least common multiply on \[\sin x\cos x\]
\[ = \dfrac{{{{\sin }^2}x\sin x + {{\sin }^2}x\cos x + \sin x{{\cos }^2}x + {{\cos }^2}x\cos x}}{{\sin x\cos x}}\]
Now we have four-part of terms in the numerator and the same denominator
And now take the common factor to the first two terms in the numerator
\[ = \dfrac{{{{\sin }^2}x(\sin x + \cos x)}}{{\sin x\cos x}}\]
And take common factor on last two terms in the numerator
\[ = \dfrac{{{{\cos }^2}x(\sin x + \cos x)}}{{\sin x\cos x}}\]
Now we reduced the four terms to two terms
\[ = \dfrac{{{{\sin }^2}x(\sin x + \cos x) + {{\cos }^2}x(\sin x + \cos x)}}{{\sin x\cos x}}\]
take common factors on both terms
\[ = \dfrac{{({{\sin }^2}x + {{\cos }^2}x)(\sin x + \cos x)}}{{\sin x\cos x}}\]
And now we use the trigonometry identity formulae \[{\sin ^2}x + {\cos ^2}x = 1\]
\[ = \dfrac{{1.(\sin x + \cos x)}}{{\sin x\cos x}}\]
\[ = \dfrac{{(\sin x + \cos x)}}{{\sin x\cos x}}\]
Now separate the terms with denominator
\[ = \dfrac{{\sin x}}{{\sin x\cos x}} + \dfrac{{\cos x}}{{\sin x\cos x}}\]

The same terms in the numerator and denominator cancel that
\[ = \dfrac{1}{{\cos x}} + \dfrac{1}{{\sin x}}\]
We use the trigonometry formulae \[\dfrac{1}{{\cos x}} = \sec x\] and \[\dfrac{1}{{\sin x}} = \operatorname{cs} cx\]
\[ = \sec x + \operatorname{cs} cx\]

Note: Important multiple terms are given to concentrate. In case we just change \[\sin x\]or \[\cos x\] the whole sum gets fully mistaken. And one more that when correct steps to only use the trigonometry formula are unnecessary don’t use that. If you unnecessarily use that trigonometry formula you can't get the proper solution quickly. Don’t get confused on the numerator and denominator terms. Because sometimes it looks the same, in that time it gives more importance.