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Question: Prove that \[\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, ...

Prove that tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = 1 + sec⁡ θ cos⁡ec θ\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\, = \,1\, + \,\sec \,\theta \,\cos ec\,\theta .

Explanation

Solution

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⁡ θ\tan \,\theta and cot⁡ θ\cot \,\theta , that is tan⁡ θ = sin⁡ θcos⁡ θ\tan \,\theta \, = \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }} and cot⁡ θ = cos⁡ θsin⁡ θ\cot \,\theta \, = \,\dfrac{{\cos \,\theta }}{{\sin \,\theta }}.
Another simplification of tan⁡ θ\tan \,\theta is tan⁡ θ = 1cot⁡ θ\tan \,\theta \, = \,\dfrac{1}{{\cot \,\theta }} and the simplification of cot⁡ θ = 1tan⁡ θ\cot \,\theta \, = \,\dfrac{1}{{\tan \,\theta }}. The formula for the reciprocal of sin⁡ θ\sin \,\theta and cos⁡ θ\cos \,\theta is 1sin⁡ θ = cos⁡ec θ\dfrac{1}{{\sin \,\theta }}\, = \,\cos ec\,\theta and 1cos⁡ θ = sec⁡ θ\dfrac{1}{{\cos \,\theta }}\, = \,\sec \,\theta .

Complete step by step solution:
Given that tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = 1 + sec⁡ θ cos⁡ec θ\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 = tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = \,\dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}
RHS =1 + sec⁡ θ cos⁡ec θ= 1\, + \,\sec \,\theta \,\cos ec\,\theta

We want to prove that LHS = RHS.
Initially solve the Left-hand side assumed part equation:
LHS =tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ  = \dfrac{{\tan \,\theta }}{{1\, - \,\cot \,\theta }}\, + \,\dfrac{{\cot \,\theta }}{{1\, - \,\tan \,\theta }}\,
By using the formula tan⁡ θ = sin⁡ θcos⁡ θ\tan \,\theta \, = \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}, substitute the formula in the above equation:
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = sin⁡ θcos⁡ θ1 − cot⁡ θ + cot⁡ θ1 − sin⁡ θcos⁡ θ\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⁡ θ\tan \,\theta , substitute the value of cot⁡ θ\cot \,\theta as cot⁡ θ = cos⁡ θsin⁡ θ\cot \,\theta \, = \,\dfrac{{\cos \,\theta }}{{\sin \,\theta }}
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = sin⁡ θcos⁡ θ1 − cos⁡ θsin⁡ θ + cos⁡ θsin⁡ θ1 − sin⁡ θcos⁡ θ\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 − cos⁡ θsin⁡ θ 1\, - \,\dfrac{{\cos \,\theta }}{{\sin \,\theta }}\,,
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = sin⁡ θcos⁡ θsin⁡ θ − cos⁡ θsin⁡ θ + cos⁡ θsin⁡ θ1 − sin⁡ θcos⁡ θ\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 − sin⁡ θcos⁡ θ1\, - \,\dfrac{{\sin \,\theta }}{{\cos \,\theta }}
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = sin⁡ θcos⁡ θsin⁡ θ − cos⁡ θsin⁡ θ + cos⁡ θsin⁡ θcos⁡ θ − sin⁡ θcos⁡ θ\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 sin⁡ θ − cos⁡ θsin⁡ θ\dfrac{{\sin \,\theta \, - \,\cos \,\theta }}{{\sin \,\theta }} becomes the reciprocal of the denominator, when it comes to nominator:
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = sin⁡ θcos⁡ θ × sin⁡ θsin⁡ θ − cos⁡ θ + cos⁡ θsin⁡ θ × cos⁡ θcos⁡ θ − sin⁡ θ\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,
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ =  sin⁡2θcos⁡ θ(sin⁡ θ − cos⁡ θ) +   cos⁡ 2θsin⁡ θ(cos⁡ θ − sin⁡ θ)\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⁡ θ − sin⁡ θ\cos \,\theta \, - \,\sin \,\theta into sin⁡ θ − cos⁡ θ\sin \,\theta \, - \,\cos \,\theta by adding the negative sign in it:

The equation becomes as follows:
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ =  sin⁡2θcos⁡ θ(sin⁡ θ − cos⁡ θ) − cos⁡ 2θsin⁡ θ(sin⁡ θ − cos⁡ θ)\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:
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = 1sin⁡ θ − cos⁡ θ (sin⁡2θcos⁡ θ − cos⁡ 2θsin⁡ θ)\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:
tan⁡ θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = 1sin⁡ θ − cos⁡ θ (sin⁡3θ −  cos⁡3 θcos⁡ θ sin⁡ θ)\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⁡3θ − cos⁡3θ{\sin ^3}\theta \, - \,{\cos ^3}\theta to the above equation.

The formula is sin⁡3θ − cos⁡3θ = (sin⁡θ − cos⁡ θ)(sin⁡2θ+cos⁡2θ+sin⁡θcos⁡θ)cos⁡θ sin⁡θ{\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 }}:
tan⁡θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = 1sin⁡ θ − cos⁡ θ (sin⁡θ − cos⁡ θ)(sin⁡2θ+cos⁡2θ+sin⁡θcos⁡θ)cos⁡θ sin⁡θ\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⁡2θ+cos⁡2θ{\sin ^2}\theta + {\cos ^2}\theta as 11 in above equation:
tan⁡θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ = 1sin⁡ θ − cos⁡ θ (sin⁡θ − cos⁡ θ)(1+sin⁡θcos⁡θ)cos⁡θ sin⁡θ\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:
tan⁡θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ =(1+sin⁡θcos⁡θ)cos⁡θ sin⁡θ\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:
tan⁡θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ =1cos⁡ θsin⁡ θ+sin⁡θcos⁡θcos⁡θ sin⁡θ\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:
tan⁡θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ =1cos⁡ θsin⁡ θ + 1\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:
tan⁡θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ =1cos⁡ θ 1sin⁡ θ + 1\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⁡θ\cos \theta and sin⁡θ\sin \theta in above equation as follows:

Since 1sin⁡ θ = cos⁡ec θ  1cos⁡ θ = sec⁡ θ\dfrac{1}{{\sin \,\theta }}\, = \,\cos ec\,\theta \,\,\dfrac{1}{{\cos \,\theta }}\, = \,\sec \,\theta in the equation:
tan⁡θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ=sec⁡θcos⁡ecθ + 1\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.
tan⁡θ1 − cot⁡ θ + cot⁡ θ1 − tan⁡ θ=cos⁡ec θsec⁡ θ + 1\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⁡θ − sin⁡θ)(\cos \theta \, - \,\sin \theta )will become − (sin⁡θ − cos⁡ θ) - \,(\sin \theta \, - \,\cos \,\theta ). Don’t forget to put the negative sign.