Solveeit Logo
Login

Question

Question: How do you prove \[\dfrac{\tan x+1}{\tan x-1}=\dfrac{\sec x+\csc x}{\sec x-\csc x}\]?...

How do you prove tan⁡x+1tan⁡x−1=sec⁡x+csc⁡xsec⁡x−csc⁡x\dfrac{\tan x+1}{\tan x-1}=\dfrac{\sec x+\csc x}{\sec x-\csc x}?

Explanation

Solution

Consider the L.H.S. of the given expression and use the conversion formula: - tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} to simplify the numerator and the denominator. Now, divide both the numerator and the denominator of the simplified L.H.S. by sin⁡xcos⁡x\sin x\cos x. Cancel the common factors and use the conversions: - 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x and 1sin⁡x=csc⁡x\dfrac{1}{\sin x}=\csc x to deduce the expression in the R.H.S.

Complete step by step answer:
Here, we have been asked to prove the relation: - tan⁡x+1tan⁡x−1=sec⁡x+csc⁡xsec⁡x−csc⁡x\dfrac{\tan x+1}{\tan x-1}=\dfrac{\sec x+\csc x}{\sec x-\csc x}. Let us start with the expression in the L.H.S. and we will try to get the expression in the R.H.S., so we have,
⇒L.H.S.=tan⁡x+1tan⁡x−1\Rightarrow L.H.S.=\dfrac{\tan x+1}{\tan x-1}
Using the conversion formula: - tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}, we get,
⇒L.H.S.=sin⁡xcos⁡x+1sin⁡xcos⁡x−1\Rightarrow L.H.S.=\dfrac{\dfrac{\sin x}{\cos x}+1}{\dfrac{\sin x}{\cos x}-1}
Taking L.C.M. and simplifying, we get,
⇒L.H.S.=sin⁡x+cos⁡xcos⁡xsin⁡x−cos⁡xcos⁡x\Rightarrow L.H.S.=\dfrac{\dfrac{\sin x+\cos x}{\cos x}}{\dfrac{\sin x-\cos x}{\cos x}}
Cancelling the denominator part, i.e., cos⁡x\cos x, from both the upper and lower part of the fraction, we get,
⇒L.H.S.=sin⁡x+cos⁡xsin⁡x−cos⁡x\Rightarrow L.H.S.=\dfrac{\sin x+\cos x}{\sin x-\cos x}
Dividing both the numerator and the denominator of the above expression by sin⁡xcos⁡x\sin x\cos x, we get,
⇒L.H.S.=sin⁡x+cos⁡xsin⁡xcos⁡xsin⁡x−cos⁡xsin⁡xcos⁡x\Rightarrow L.H.S.=\dfrac{\dfrac{\sin x+\cos x}{\sin x\cos x}}{\dfrac{\sin x-\cos x}{\sin x\cos x}}
Breaking the terms, we get,
⇒L.H.S.=sin⁡xsin⁡xcos⁡x+cos⁡xsin⁡xcos⁡xsin⁡xsin⁡xcos⁡x−cos⁡xsin⁡xcos⁡x\Rightarrow L.H.S.=\dfrac{\dfrac{\sin x}{\sin x\cos x}+\dfrac{\cos x}{\sin x\cos x}}{\dfrac{\sin x}{\sin x\cos x}-\dfrac{\cos x}{\sin x\cos x}}
Cancelling the common factors, we get,
⇒L.H.S.=1cos⁡x+1sin⁡x1cos⁡x−1sin⁡x\Rightarrow L.H.S.=\dfrac{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}{\dfrac{1}{\cos x}-\dfrac{1}{\sin x}}
Using the conversions: - 1sin⁡x=csc⁡x\dfrac{1}{\sin x}=\csc x and 1cos⁡x=sec⁡x\dfrac{1}{\cos x}=\sec x, we get,
⇒L.H.S.=sec⁡x+csc⁡xsec⁡x−csc⁡x\Rightarrow L.H.S.=\dfrac{\sec x+\csc x}{\sec x-\csc x}
⇒L.H.S.=R.H.S.\Rightarrow L.H.S.=R.H.S.
Hence, proved

Note:
One may note that here we have started with the expression in the L.H.S. and deduced in the expression in the R.H.S. You can also use the reverse process, i.e., start with the expression in the R.H.S. and reach the expression in the L.H.S. There can be a third way also to prove the relation. What we can do is we will start with the L.H.S. and simplify it to a particular step and then start solving the R.H.S. and try to get the simplified expression at which we stopped solving the L.H.S. You must remember all the conversion formulas that relate the six trigonometric functions with each other.