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# How do you verify $\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=2\cot x\csc x$ ?

Last updated date: 14th Jun 2024
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Hint: First analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation. This can be done by using the suitable trigonometric identities (formulae) and simplify any one side of the equation.

Complete step by step solution:
Sine, cosine and tangent of an angle are trigonometric ratios. These ratios are also called trigonometric functions. When we plot a graph of the trigonometric ratios with respect to all the real values of an angle, we get a graph that has a periodic property. This means that the graph repeats itself after equal intervals of the angleOther than the trigonometric ratios sine, cosine and tangent we have other trigonometric ratios called cosecant, secant and cotangent.

All the above six trigonometric ratios (functions) are dependent on each other. There are different properties and identities that relate the trigonometric ratios.The equation that has to be verified is,
$\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=2\cot x\csc x$
Let us first analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation.The left hand side of the equation is,
$\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}$ ….. (i)
Let us first make a common denominator in the above expression.With this we get that,
$\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=\dfrac{(\sec x+1)+(\sec x-1)}{(\sec x-1)(\sec x+1)}$
$\Rightarrow \dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=\dfrac{2\sec x}{{{\sec }^{2}}x-1}$ …. (ii)
We know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$.
Then gives us that ${{\sec }^{2}}x-1={{\tan }^{2}}x$

Substitute this value in (ii)
$\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=\dfrac{2\sec x}{{{\tan }^{2}}x}$
Now, substitute $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}$ in the right hand side.Then,
$\Rightarrow \dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=\dfrac{\dfrac{2}{\cos x}}{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}=\dfrac{2\cos x}{{{\sin }^{2}}x}$.
We can write the above equation as,
$\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=\dfrac{2}{\sin x\tan x}$
Now, substitute $\csc x=\dfrac{1}{\sin x}$ and $\cot x=\dfrac{\cos x}{\sin x}$ in the right hand side.
$\therefore\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=2\cot x\csc x$

Therefore, the left hand side of the equation is equal to the right hand side of the equation.Hence, the given equation is correct.

Note: It is not compulsory to verify a given equation only by simplifying the left hand side of the equation. You can also simply the right hand side and check whether it results as the same as the left hand side. You may also analyse both the sides of the given equation and simplify them in terms of sine and cosine functions. If both the simplifications match then the given equation is correct.