Question

The value of $x$ from the equation $\text{cosec}\left( {{90}^{\circ }}-\theta \right)-x\sin \left( {{90}^{\circ }}-\theta \right)\tan \left( {{180}^{\circ }}+\theta \right)=\sin \left( {{90}^{\circ }}+\theta \right)$ is
(a) $\sin \theta$
(b) $\cos \theta$
(c) $\tan \theta$
(d) $\sec \theta$

Answer 1

We must know the transformations that take place ${{90}^{\circ }}\text{ or 18}{{0}^{\circ }}$ is added or subtracted from the angle. These formulae are $\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta$, $\text{cosec}\left( {{90}^{\circ }}-\theta \right)=\sec \theta$, $\sin \left( {{90}^{\circ }}+\theta \right)=\cos \theta$ and $\tan \left( {{180}^{\circ }}+\theta \right)=\tan \theta$. We must also know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find the value of $x$.

Complete step-by-step solution:
We are given the following trigonometric equation,
$\text{cosec}\left( {{90}^{\circ }}-\theta \right)-x\sin \left( {{90}^{\circ }}-\theta \right)\tan \left( {{180}^{\circ }}+\theta \right)=\sin \left( {{90}^{\circ }}+\theta \right)$.
We know that the trigonometric ratios of $\theta$ and ${{90}^{\circ }}-\theta$ are related as
$\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta$
$\text{cosec}\left( {{90}^{\circ }}-\theta \right)=\sec \theta$
Using the above two values, we can transform the given equation into the following form,
$\sec \theta -x\cos \theta \tan \left( {{180}^{\circ }}+\theta \right)=\sin \left( {{90}^{\circ }}+\theta \right)$
We also know that the relation between the trigonometric ratios in $\theta$ and ${{90}^{\circ }}+\theta$ is
$\sin \left( {{90}^{\circ }}+\theta \right)=\cos \theta$
Hence, we can write
$\sec \theta -x\cos \theta \tan \left( {{180}^{\circ }}+\theta \right)=\cos \theta$
Also, we all know very well that the relation between trigonometric ratios in $\theta$ and ${{180}^{\circ }}+\theta$ is
$\tan \left( {{180}^{\circ }}+\theta \right)=\tan \theta$
Thus, our equation now becomes
$\sec \theta -x\cos \theta \tan \theta =\cos \theta$
It is known to us that the trigonometric ratio tangent is defined as $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.
Thus, we now have
$\sec \theta -x\cos \theta \dfrac{\sin \theta }{\cos \theta }=\cos \theta$
On cancelling the cosine term, we get the following simplified equation,
$\sec \theta -x\sin \theta =\cos \theta$
We can rearrange these terms to get the following equation,
$x\sin \theta =\sec \theta -\cos \theta$
We know that the secant of an angle is nothing, but the reciprocal of cosine of the same angle. We can write this mathematically as $\sec \theta =\dfrac{1}{\cos \theta }$.
On replacing the value, we get
$x\sin \theta =\dfrac{1}{\cos \theta }-\cos \theta$
Let us take the term $\cos \theta$ as LCM on the right hand side. Thus, we get
$x\sin \theta =\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta }...\left( i \right)$
We all know very well that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. We can rewrite this property as $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta$.
Using the above property, we can write equation (i) as
$x\sin \theta =\dfrac{{{\sin }^{2}}\theta }{\cos \theta }$
Let us now cancel the term $\sin \theta$ from both sides of the above equation. Thus, we get
$x=\dfrac{\sin \theta }{\cos \theta }$
By using the above definition of $\tan \theta$, we can easily write
$x=\tan \theta$.
Hence, option (c) is the correct answer.

Note: We can remember all the transformations for trigonometric identities using the acronym ASTC or All Silver Tea Cups. Here, A stands for all positive in the first quadrant, S stands for sin (and cosec) positive in the second quadrant, T stands for tan (and cot) positive in the third quadrant and C stands for cos (and sec) positive in the fourth quadrant.