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Question: If \[\tan \theta = \dfrac{{x\sin \varphi }}{{1 - x\cos \varphi }}\] and \[\tan \varphi = \dfrac{{y\s...

If tan⁡θ=xsin⁡φ1−xcos⁡φ\tan \theta = \dfrac{{x\sin \varphi }}{{1 - x\cos \varphi }} and tan⁡φ=ysin⁡θ1−ycos⁡θ\tan \varphi = \dfrac{{y\sin \theta }}{{1 - y\cos \theta }} then xy\dfrac{x}{y} is equal to
(A) sin⁡φsin⁡θ\dfrac{{\sin \varphi }}{{\sin \theta }}
(B) sin⁡θsin⁡φ\dfrac{{\sin \theta }}{{\sin \varphi }}
(C) sin⁡θ1−cos⁡θ\dfrac{{\sin \theta }}{{1 - \cos \theta }}
(D) sin⁡θ1−cos⁡φ\dfrac{{\sin \theta }}{{1 - \cos \varphi }}

Explanation

Solution

First of all, do the inverse of both equation and the solve equation by breaking the terms in numerator and then substituting the values of trigonometric terms in fractions as other terms and shifting all the values to one side and the variable x or y to other side to form two separate equations. Then we take the ratio of that equation and find the value of xy\dfrac{x}{y}.
Inverse of an element is that value of an element which when taken reciprocal of gives the element itself.

Complete step-by-step answer :
Let’s solve, the
tan⁡θ=xsin⁡φ1−xcos⁡φ\tan \theta = \dfrac{{x\sin \varphi }}{{1 - x\cos \varphi }}
In above equation do the inverse on both side,
\Rightarrow$$$\cot \theta = \dfrac{{1 - x\cos \varphi }}{{x\sin \varphi }}$$ Now separate the terms in the numerator i.e.1 and$$x\cos \varphi $$. \Rightarrow\cot \theta = \dfrac{1}{{x\sin \varphi }} - \dfrac{{x\cos \varphi }}{{x\sin \varphi }}$$ Put the value of $$\cot \varphi = \dfrac{{\cos \varphi }}{{\sin \varphi }}$$ $\Rightarrow\cot \theta = \dfrac{1}{{x\sin \varphi }} - \cot \varphi $\Rightarrow$$$\cot \theta + \cot \varphi = \dfrac{1}{{x\sin \varphi }} (Equation 1)
Now, take
tan⁡φ=ysin⁡θ1−ycos⁡θ\tan \varphi = \dfrac{{y\sin \theta }}{{1 - y\cos \theta }}
In above equation do the inverse on both side,
\Rightarrow$$$\cot \varphi = \dfrac{{1 - y\cos \theta }}{{y\sin \theta }}$$ Now separate the terms in the numerator i.e. 1 and$$y\cos \theta $$. \Rightarrow\cot \varphi = \dfrac{1}{{y\sin \theta }} - \dfrac{{y\cos \theta }}{{y\sin \theta }}$$ Cancel out the same terms from numerator and denominator. $\Rightarrow\cot \varphi = \dfrac{1}{{y\sin \theta }} - \cot \theta Shift the same trigonometric values to one side of the equation. $\Rightarrow$$$\cot \theta + \cot \varphi = \dfrac{1}{{y\sin \theta }} (Equation 2)
Now divide equation 1 to equation 2 we get,
\Rightarrow$$$\dfrac{{\cot \theta + \cot \varphi }}{{\cot \theta + \cot \varphi }} = \dfrac{{\dfrac{1}{{x\sin \varphi }}}}{{\dfrac{1}{{y\sin \theta }}}}$$ Cancel out the same terms from the numerator and denominator in LHS of the equation. \Rightarrow1 = \dfrac{{y\sin \theta }}{{x\sin \varphi }}$$ Shift the variables to left hand side of the equation $\Rightarrow\dfrac{x}{y} = \dfrac{{\sin \theta }}{{\sin \varphi }}$$

∴\therefore The value of xy\dfrac{x}{y} is sin⁡θsin⁡φ\dfrac{{\sin \theta }}{{\sin \varphi }}. Therefore, option A is correct.

Note :
In the beginning of the question we take inverse on both sides so don’t forget to take the inverse of tan⁡θ\tan \theta .
And the inverse of tan⁡θ\tan \theta is cot⁡θ\cot \theta . Some students forgot to do the inverse and got errors in answers.
Students can transform the trigonometric terms by dividing or multiplying the terms with any other term so as to make the solution easy.