Question

How do you prove $ \dfrac{\sec x}{\sin x}-\dfrac{\sin x}{\cos x}=\cot x $ ?

Answer 1

Hint: To prove the above trigonometric equation i.e. $ \dfrac{\sec x}{\sin x}-\dfrac{\sin x}{\cos x}=\cot x $ , we are going to write $ \sec x=\dfrac{1}{\cos x} $ in this equation and then will take the L.C.M of denominator on the L.H.S of this equation. Further simplification of L.H.S will make the L.H.S equal to R.H.S.

Complete step by step answer:
The trigonometric equation given in the above problem is that:
 $ \dfrac{\sec x}{\sin x}-\dfrac{\sin x}{\cos x}=\cot x $
Now, for proving the above equation, we are going to simplify L.H.S of the above equation. For that we are going to write $ \sec x=\dfrac{1}{\cos x} $ in the L.H.S and we get,
 $ \Rightarrow \dfrac{1}{\sin x\cos x}-\dfrac{\sin x}{\cos x}=\cot x $
Taking L.C.M of the denominator in the L.H.S of the above equation we get the L.C.M of the denominator as $ \sin x\cos x $ .
 $ \Rightarrow \dfrac{1-\sin x\left( \sin x \right)}{\sin x\cos x}=\cot x $
Multiplying $ \sin x $ by $ \sin x $ in the numerator of the L.H.S of the above equation we get,
 $ \Rightarrow \dfrac{1-{{\sin }^{2}}x}{\sin x\cos x}=\cot x $
Now, we know the trigonometric identity that:
 $ 1-{{\sin }^{2}}x={{\cos }^{2}}x $
Using the above identity in $ \dfrac{1-{{\sin }^{2}}x}{\sin x\cos x}=\cot x $ we get,
 $ \Rightarrow \dfrac{{{\cos }^{2}}x}{\sin x\cos x}=\cot x $
From the numerator and denominator of the L.H.S of the above equation, $ \cos x $ will be cancelled out and we are left with:
 $ \Rightarrow \dfrac{\cos x}{\sin x}=\cot x $
Also, there is a trigonometric identity that $ \dfrac{\cos x}{\sin x}=\cot x $ so using this identity in the L.H.S of the above equation we get,
 $ \Rightarrow \cot x=\cot x $
As we have shown that L.H.S = R.H.S so we have proved the given equation.

Note:
 In the above solution, after this step $ \dfrac{{{\cos }^{2}}x}{\sin x\cos x}=\cot x $ , you can change the way of approaching this problem. You can write $ \cot x=\dfrac{\cos x}{\sin x} $ and then this equation will look like:
 $ \Rightarrow \dfrac{{{\cos }^{2}}x}{\sin x\cos x}=\dfrac{\cos x}{\sin x} $
Now, $ \sin x $ will be cancelled out in the denominator of both the sides we get,
 $ \Rightarrow \dfrac{{{\cos }^{2}}x}{\cos x}=\cos x $
From the numerator and denominator of the L.H.S of the above equation, $ \cos x $ will be cancelled out and we are left with:
 $ \Rightarrow \cos x=\cos x $
As you can see that L.H.S = R.H.S in the above equation so we have proved the given equation in some other way.