Question

How do you write $\tan \theta -\cot \theta $ in terms of $\cos \theta $ ?

Answer 1

Hint: We can write any trigonometry equation in the terms of any trigonometric identity. We know the relation between all 6 identities. sin x and cosec x , cos x and sec x , tan x and cot x are reciprocal of each other. ${{\sin }^{2}}x+{{\cos }^{2}}=1$ , tan x is the ratio of sin x and cos x, cot x is the ratio of cos x and sin x.

Complete step-by-step answer:
We have to write $\tan \theta -\cot \theta $ in term of $\cos \theta $
We know that tan x is the ratio of sin x and cos x, cot x is the ratio of cos x and sin x.
So we can write $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
So we can write $\tan \theta -\cot \theta $ = $\dfrac{\sin \theta }{\cos \theta }-\dfrac{\cos \theta }{\sin \theta }$
Solving the fraction we get $\tan \theta -\cot \theta =\dfrac{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }{\sin \theta \cos \theta }$
We can write ${{\sin }^{2}}x=1-{{\cos }^{2}}x$
So $\tan \theta -\cot \theta =\dfrac{1-2{{\cos }^{2}}\theta }{\sin \theta \cos \theta }$
We can write $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ when $\sin \theta $ is positive and $\sin \theta $ is equal to $-\sqrt{1-{{\cos }^{2}}\theta }$ when $\sin \theta $ is negative.
So we can write $\tan \theta -\cot \theta =\dfrac{1-2{{\cos }^{2}}\theta }{\sqrt{1-{{\cos }^{2}}\theta }\cos \theta }$ ; $\theta \in \left( 2n\pi ,2n+1\pi \right)$ - all the angle where cos is either 0 , n is an integer
    $\tan \theta -\cot \theta =-\dfrac{1-2{{\cos }^{2}}\theta }{\sqrt{1-{{\cos }^{2}}\theta }\cos \theta }$ ; $\theta \in \left( 2n-1\pi ,2n\pi \right)$ - all the angles where cos is 0, n is an integer.

Note: In the answer to the above question we have excluded all the angles at which cos x is equal to 0, because if we put cos x as 0 then the denominator will become 0. If cos x will be 1 then also the denominator will be 0, for that we already did not have any angle $n\pi $ in the range.