Question

Find x from the following equation.
\[\begin{align}
  & \left( 1 \right)\text{cosec}\left( \dfrac{\pi }{2}+\theta \right)+x\cos \theta \cot \left( \dfrac{\pi }{2}+\theta \right)=\sin \left( \dfrac{\pi }{2}+\theta \right) \\
 & \left( 2 \right)x\cot \left( \dfrac{\pi }{2}+\theta \right)+\tan \left( \dfrac{\pi }{2}+\theta \right)\sin \theta +\text{cosec}\left( \dfrac{\pi }{2}+\theta \right)=0 \\
\end{align}\]

Answer 1

Hint: In this question, we are given two-equation and we need to find the values of x from them. For this, we will use a various trigonometric formula which is as follows:
(i) $ \left( \dfrac{\pi }{2}+\theta \right) $ lies in second quadrant where only sine and cosecant are positive.
\[\begin{align}
  & \left( ii \right)\tan \left( \dfrac{\pi }{2}+\theta \right)=-\cot \theta \\
 & \left( iii \right)\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta \\
 & \left( iv \right)\text{cosec}\left( \dfrac{\pi }{2}+\theta \right)=\sec \theta \\
 & \left( v \right)\sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta \\
 & \left( vi \right)\sec \theta =\dfrac{1}{\cos \theta } \\
 & \left( vii \right)\cot \theta =\dfrac{\cos \theta }{\sin \theta } \\
 & \left( viii \right)\tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \left( ix \right){{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
\end{align}\]

Complete step by step answer:
Let us solve part (1)
Here we are given the equation as,
\[\text{cosec}\left( \dfrac{\pi }{2}+\theta \right)+x\cos \theta \cot \left( \dfrac{\pi }{2}+\theta \right)=\sin \left( \dfrac{\pi }{2}+\theta \right)\].
We need to find the value of x.
We know that $ \left( \dfrac{\pi }{2}+\theta \right) $ lies in the second quadrant where only sine and cosecant are positive.
So we can say that, $ \sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta ,\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta \text{ and cosec}\left( \dfrac{\pi }{2}+\theta \right)=\sec \theta $ .
Putting all these values in the above equation we get,
 $ \sec \theta +x\cos \theta \left( -\tan \theta \right)=\cos \theta $ .
Simplifying we get,
 $ \sec \theta -x\cos \theta \tan \theta =\cos \theta $ .
We know that $ \sec \theta $ is reciprocal of $ \cos \theta $ so $ \sec \theta =\dfrac{1}{\cos \theta }\text{ and }\tan \theta =\dfrac{\sin \theta }{\cos \theta } $ so we get,
 $ \dfrac{1}{\cos \theta }-x\cos \theta \cdot \dfrac{\sin \theta }{\cos \theta }=\cos \theta $ .
Cancelling $ \cos \theta $ from the numerator and the denominator in the second term we get,
 $ \dfrac{1}{\cos \theta }-x\sin \theta =\cos \theta $ .
Rearranging the terms we get,
 $ x\sin \theta =\dfrac{1}{\cos \theta }-\cos \theta $ .
Taking LCM of $ \cos \theta $ in the right side we get,
 $ x\sin \theta =\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } $ .
We know that, $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ so we can write it as $ {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ . Putting it in above equation we get $ x\sin \theta =\dfrac{{{\sin }^{2}}\theta }{\cos \theta } $ .
Cancelling $ \sin \theta $ from both sides we get,
 $ x=\dfrac{\sin \theta }{\cos \theta } $ .
 $ \dfrac{\sin \theta }{\cos \theta } $ can be written as $ \tan \theta $ so we get, $ x=\tan \theta $ which is our required answer.
(2) \[x\cot \left( \dfrac{\pi }{2}+\theta \right)+\tan \left( \dfrac{\pi }{2}+\theta \right)\sin \theta +\text{cosec}\left( \dfrac{\pi }{2}+\theta \right)=0\].
Using the formula of $ \tan \left( \dfrac{\pi }{2}+\theta \right)=-\cot \theta ,\cot \left( \dfrac{\pi }{2}+\theta \right)=-\tan \theta \text{ and cosec}\left( \dfrac{\pi }{2}+\theta \right)=\sec \theta $ we get,
 $ x\left( -\tan \theta \right)+\left( -\cot \theta \right)\sin \theta +\sec \theta =0 $ .
Simplifying we get,
 $ -x\tan \theta -\cot \theta \sin \theta +\sec \theta =0 $ .
We know that $ \tan \theta $ is equal to $ \dfrac{\sin \theta }{\cos \theta }\text{ and cot}\theta =\dfrac{\cos \theta }{\sin \theta } $ . Also $ \sec \theta $ is reciprocal of $ \cos \theta $ so, $ \sec \theta =\dfrac{1}{\cos \theta } $ . Putting all these values in the above equation we get,
 $ -x\dfrac{\sin \theta }{\cos \theta }-\dfrac{\cos \theta }{\sin \theta }\sin \theta +\dfrac{1}{\cos \theta }=0 $ .
Cancelling $ \sin \theta $ from the numerator and the denominator in the second term we get,
 $ -x\dfrac{\sin \theta }{\cos \theta }-\cos \theta +\dfrac{1}{\cos \theta }=0 $ .
Rearranging we get,
 $ x\dfrac{\sin \theta }{\cos \theta }=\dfrac{1}{\cos \theta }-\cos \theta $ .
Taking LCM of $ \cos \theta $ in the right side we get,
 $ x\dfrac{\sin \theta }{\cos \theta }=\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } $ .
We know that $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ so we can write it as $ {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ . Putting it in the above equation we get $ x\dfrac{\sin \theta }{\cos \theta }=\dfrac{{{\sin }^{2}}\theta }{\cos \theta } $ .
Cancelling $ \dfrac{\sin \theta }{\cos \theta } $ from both sides we get, $ x=\sin \theta $ which is our required answer.

Note:
Students should keep in mind all the trigonometric formulas before solving this sum. Take care of signs while solving the equations. Students can check their answer by putting the value of x in the original equation and checking if the left side is equal to the right side.