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

Last updated date: 05th Mar 2024
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IVSAT 2024
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Hint: We first take the left-hand part of the equation of $\sec x-\cos x=\dfrac{\sin x}{\cot x}$. Then we simplify the equation. We convert the denominator using the relation $\sec x=\dfrac{1}{\cos x}$. Then we use the theorem \[{{\cos }^{2}}x+{{\sin }^{2}}x=1\]. We eliminate the part \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\]. We then divide $\sin x$ from both denominator and numerator After elimination we get the right-hand side of the equation.

Complete step-by-step solution:
We have to prove the trigonometric equation $\sec x-\cos x=\dfrac{\sin x}{\cot x}$.
We take the left-hand side of the equation $\sec x-\cos x=\dfrac{\sin x}{\cot x}$ and prove the right-side part.
We get $\sec x-\cos x$. We know that $\sec x=\dfrac{1}{\cos x}$.
Therefore, $\sec x-\cos x=\dfrac{1}{\cos x}-\cos x=\dfrac{1-{{\cos }^{2}}x}{\cos x}$.
We now use the identity theorem of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ which gives us ${{\sin }^{2}}x=1-{{\cos }^{2}}x$. We place the value in the equation and get $\dfrac{1-{{\cos }^{2}}x}{\cos x}=\dfrac{{{\sin }^{2}}x}{\cos x}$.
We can now divide $\sin x$ from both denominator and numerator.
The equation becomes $\dfrac{{}^{{{\sin }^{2}}x}/{}_{\sin x}}{{}^{\cos x}/{}_{\sin x}}$.
Now we apply the theorem \[\cot x=\dfrac{\cos x}{\sin x}\] again to convert to $\cot x$.
The final form is $\dfrac{{}^{{{\sin }^{2}}x}/{}_{\sin x}}{{}^{\cos x}/{}_{\sin x}}=\dfrac{\sin x}{\cot x}$.
Thus proved $\sec x-\cos x=\dfrac{\sin x}{\cot x}$.

Note: It is important to remember that the condition to eliminate the $\sin x$ from both denominator and numerator is $\sin x\ne 0$. No domain is given for the variable $x$. The simplified condition will be $x\ne n\pi ,n\in \mathbb{Z}$. The identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $\sec x=\dfrac{1}{\cos x}$ are valid for any value of $x$. The division of the fraction part only gives $\sin x$ as the solution.

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