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# Prove that $\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,1\, + \,\sec \,\theta \,\cos ec\,\theta$.

Last updated date: 18th Jun 2024
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Answer
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Hint: Split the equation as LHS and RHS to prove that the given equation is true. Initially solve the left-hand side equation. By solving the left-hand side, at the final stage the assumed Right-hand side will be the answer of the solved LHS.

Useful Formula:
Use the trigonometric formula for $\tan \,\theta$and $\cot \,\theta$, that is $\tan \,\theta \, = \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}$ and $\cot \,\theta \, = \,\dfrac{{\cos \,\theta }}{{\sin \,\theta }}$.
Another simplification of $\tan \,\theta$ is $\tan \,\theta \, = \,\dfrac{1}{{\cot \,\theta }}$ and the simplification of $\cot \,\theta \, = \,\dfrac{1}{{\tan \,\theta }}$. The formula for the reciprocal of $\sin \,\theta$ and $\cos \,\theta$ is $\dfrac{1}{{\sin \,\theta }}\, = \,\cos ec\,\theta$ and $\dfrac{1}{{\cos \,\theta }}\, = \,\sec \,\theta$.

Complete step by step solution:
Given that $\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,1\, + \,\sec \,\theta \,\cos ec\,\theta$ and We want to prove it.
To prove the given equation is correct, we initially split the equation by two parts as left hand side and right-hand side.
LHS $= \,\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}$
RHS $= 1\, + \,\sec \,\theta \,\cos ec\,\theta$

We want to prove that LHS = RHS.
Initially solve the Left-hand side assumed part equation:
LHS $= \dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\,$
By using the formula $\tan \,\theta \, = \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}$, substitute the formula in the above equation:
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{{\dfrac{{\sin \,\theta }}{{\cos \,\theta }}}}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}}}$

As like $\tan \,\theta$, substitute the value of $\cot \,\theta$ as $\cot \,\theta \, = \,\dfrac{{\cos \,\theta }}{{\sin \,\theta }}$
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{{\dfrac{{\sin \,\theta }}{{\cos \,\theta }}}}{{1\, - \,\dfrac{{\cos \,\theta }}{{\sin \,\theta }}}}\, + \,\dfrac{{\dfrac{{\cos \,\theta }}{{\sin \,\theta }}}}{{1\, - \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}}}$
Taking LCM to $1\, - \,\dfrac{{\cos \,\theta }}{{\sin \,\theta }}\,$,
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{{\dfrac{{\sin \,\theta }}{{\cos \,\theta }}}}{{\dfrac{{\sin \,\theta \, - \,\cos \,\theta }}{{\sin \,\theta }}}}\, + \,\dfrac{{\dfrac{{\cos \,\theta }}{{\sin \,\theta }}}}{{1\, - \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}}}$
Similar that, take LCM to $1\, - \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}$
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{{\dfrac{{\sin \,\theta }}{{\cos \,\theta }}}}{{\dfrac{{\sin \,\theta \, - \,\cos \,\theta }}{{\sin \,\theta }}}}\, + \,\dfrac{{\dfrac{{\cos \,\theta }}{{\sin \,\theta }}}}{{\dfrac{{\cos \,\theta \, - \,\sin \,\theta }}{{\cos \,\theta }}}}$

The denominator of $\dfrac{{\sin \,\theta \, - \,\cos \,\theta }}{{\sin \,\theta }}$ becomes the reciprocal of the denominator, when it comes to nominator:
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}\, \times \,\dfrac{{\sin \,\theta }}{{\sin \,\theta \, - \,\cos \,\theta }}\, + \,\dfrac{{\cos \,\theta }}{{\sin \,\theta }}\, \times \,\dfrac{{\cos \,\theta }}{{\cos \,\theta \, - \,\sin \,\theta }}$
To simplify, multiply the equation as possible,
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\,\dfrac{{{{\sin }^2}\theta }}{{\cos \,\theta (\sin \,\theta \, - \,\cos \,\theta )}}\, + \,\,\,\dfrac{{\cos {\,^2}\theta }}{{\sin \,\theta (\cos \,\theta \, - \,\sin \,\theta )}}$
Change the $\cos \,\theta \, - \,\sin \,\theta$into $\sin \,\theta \, - \,\cos \,\theta$by adding the negative sign in it:

The equation becomes as follows:
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\,\dfrac{{{{\sin }^2}\theta }}{{\cos \,\theta (\sin \,\theta \, - \,\cos \,\theta )}}\, - \,\dfrac{{\cos {\,^2}\theta }}{{\sin \,\theta (\sin \,\theta \, - \,\cos \,\theta )}}$
By taking out the common parts from the equation as follows:
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{1}{{\sin \,\theta \, - \,\cos \,\theta }}\,(\dfrac{{{{\sin }^2}\theta }}{{\cos \,\theta }}\, - \,\dfrac{{\cos {\,^2}\theta }}{{\sin \,\theta }})$
Take the LCM to the above equation except common equation:
The equation should be as follows:
$\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{1}{{\sin \,\theta \, - \,\cos \,\theta }}\,(\dfrac{{{{\sin }^3}\theta \, - \,\,{{\cos }^3}\,\theta }}{{\cos \,\theta \,\sin \,\theta }})$
Substitute the formula for ${\sin ^3}\theta \, - \,{\cos ^3}\theta$ to the above equation.

The formula is ${\sin ^3}\theta \, - \,{\cos ^3}\theta \, = \,\dfrac{{(\sin \theta \, - \,\cos \,\theta )({{\sin }^2}\theta + {{\cos }^2}\theta + \sin \theta \cos \theta )}}{{\cos \theta \,\sin \theta }}$:
$\dfrac{{\tan \theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{1}{{\sin \,\theta \, - \,\cos \,\theta }}\,\dfrac{{(\sin \theta \, - \,\cos \,\theta )({{\sin }^2}\theta + {{\cos }^2}\theta + \sin \theta \cos \theta )}}{{\cos \theta \,\sin \theta }}$
Substitute the value of ${\sin ^2}\theta + {\cos ^2}\theta$ as $1$ in above equation:
$\dfrac{{\tan \theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,\dfrac{1}{{\sin \,\theta \, - \,\cos \,\theta }}\,\dfrac{{(\sin \theta \, - \,\cos \,\theta )(1 + \sin \theta \cos \theta )}}{{\cos \theta \,\sin \theta }}$
Now cancel the common parts from both numerator and denominator, then the equation become as follows:
$\dfrac{{\tan \theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \dfrac{{(1 + \sin \theta \cos \theta )}}{{\cos \theta \,\sin \theta }}$

Splitting the above equation in two parts:
$\dfrac{{\tan \theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \dfrac{1}{{\cos \,\theta \sin \,\theta }} + \dfrac{{\sin \theta \cos \theta }}{{\cos \theta \,\sin \theta }}$
Cancel the common parts from both numerator and denominator in above equation as follows:
$\dfrac{{\tan \theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \dfrac{1}{{\cos \,\theta \sin \,\theta }}\, + \,1$

Splitting the right-hand side equation in again two parts as follows:
$\dfrac{{\tan \theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \dfrac{1}{{\cos \,\theta }}\,\dfrac{1}{{\sin \,\theta }}\, + \,1$
Now substitute the value of reciprocal value of $\cos \theta$and $\sin \theta$in above equation as follows:

Since $\dfrac{1}{{\sin \,\theta }}\, = \,\cos ec\,\theta \,\,\dfrac{1}{{\cos \,\theta }}\, = \,\sec \,\theta$in the equation:
$\dfrac{{\tan \theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }} = \sec \theta \cos ec\theta \, + \,1$
By simplifying the Left-hand side, we obtain the Right-hand side.

We proved that the given equation is correct.
$\dfrac{{\tan \theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }} = \cos ec\,\theta \sec \,\theta \, + \,1$ is proved.

Note: Only take the LCM to the applicable places and use the formula only in needed places. Be aware that changing the equation like $(\cos \theta \, - \,\sin \theta )$will become $- \,(\sin \theta \, - \,\cos \,\theta )$. Don’t forget to put the negative sign.