Question

How do you prove $\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}} = 1$?

Answer 1

Hint: This question is related to the trigonometry, and we have to prove that the left hand side is equal to the right hand side of the expression, and this question can be solved by using trigonometric identities i.e.,${\cos ^2}A + {\sin ^2}A = 1$, and by further simplifying we will get the result which is on the right hand.

Complete step by step solution:
Given function is $\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}} = 1$,
Now we have to prove that left hand side is equal to the right hand side of the equation, now take the expression on the left hand side i.e.,
$ \Rightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}}$,
Now taking L.C.M we get,
$ \Rightarrow \dfrac{{\sin x \times \sin x}}{{\cos x \times \sin x}} + \dfrac{{\cos x \times \cos x}}{{\sin x \times \cos x}}$,
Now simplifying we get,
$ \Rightarrow \dfrac{{{{\sin }^2}x}}{{\cos x\sin x}} + \dfrac{{{{\cos }^2}x}}{{\sin x\cos x}}$ ,
Now adding the two terms we get,
$ \Rightarrow \dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x\sin x}}$
Now using the trigonometric identities i.e., ${\cos ^2}A + {\sin ^2}A = 1$ , we get,
$ \Rightarrow \dfrac{1}{{\cos x\sin x}}$,
Now, this can only be true if the denominator is equal to 1 which would mean that $\sin x = \dfrac{1}{{\cos x}}$ for all $x$. As$\cos x = \dfrac{1}{{\sec x}}$ and $\sec x$ is certainly not the same as $\sin x$, your equation can't be an identity.
Which is not equal to the right hand side of the equation,
Hence the given function is not an identity and it cannot be proved.

Note: An identity is an equation that always holds true. A trigonometric identity is an identity that contains trigonometric functions and holds true for all right-angled triangles. They are useful when solving questions with trigonometric functions and expressions. There are many trigonometric identities, here are some useful identities:
$\tan x = \dfrac{{\sin x}}{{\cos x}}$,
$\sec x = \dfrac{1}{{\cos x}}$,
$\cot x = \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}$ ,
${\sin ^2}x = 1 - {\cos ^2}x$,
${\cos ^2}x + {\sin ^2}x = 1$,
${\sec ^2}x - {\tan ^2}x = 1$,
${\csc ^2}x = 1 + {\cot ^2}x$.
${\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x$,
${\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1$,
$\sin 2x = 2\sin x\cos x$,
$2{\cos ^2}x = 1 + \cos 2x$,
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$.