
Find the value of $x$. If $\cos \left( 270{}^\circ +\theta \right)+x\cdot \cos \theta \cdot \cot \left( 180{}^\circ +\theta \right)=\sin \left( 270{}^\circ -\theta \right)$.
Answer
465.6k+ views
Hint: We must know about the relations among the different values given by sine and cosine functions on having angles that differ by multiples of $90{}^\circ $. These relations are mentioned below
$\begin{align}
& \sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \\
& \sin \left( {{90}^{\circ }}+\theta \right)=\cos \theta \\
& \sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta \\
& \sin \left( {{180}^{\circ }}+\theta \right)=-\sin \theta \\
& \cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \\
& \cos \left( {{90}^{\circ }}+\theta \right)=\sin \theta \\
& \cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta \\
& \cos \left( {{180}^{\circ }}+\theta \right)=-\cos \theta \\
\end{align}$
Now, using these expressions and relations of sine and cosine functions, we will simplify the given equation and find the value of $x$.
Complete step by step answer:
We will first simplify the L.H.S. of the given equation by using the relations mentioned above, in the following manner,
L.H.S. $=\cos \left( 270{}^\circ +\theta \right)+x\cdot \cos \theta \cdot \cot \left( 180{}^\circ +\theta \right)$
Now, we know that $\cot \left( {{180}^{\circ }}+\theta \right)=\dfrac{\cos \left( {{180}^{\circ }}+\theta \right)}{\sin \left( {{180}^{\circ }}+\theta \right)}=\dfrac{-\cos \theta }{-\sin \theta }=\dfrac{\cos \theta }{\sin \theta }$.
The value for $\cos \left( {{270}^{\circ }}+\theta \right)=\cos \left( {{180}^{\circ }}+{{90}^{\circ }}+\theta \right)=\cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)$ . We will use the relations mentioned above in the following manner,
$\cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)=-\cos \left( {{90}^{\circ }}+\theta \right)=\sin \theta $. Therefore, we have $\cos \left( {{270}^{\circ }}+\theta \right)=\sin \theta $.
Substituting these values in the L.H.S., we get
L.H.S. $=\sin \theta +x\cdot \cos \theta \cdot \dfrac{\cos \theta }{\sin \theta }$
Simplifying this equation, we get
L.H.S. $=\sin \theta +x\cdot \dfrac{{{\cos }^{2}}\theta }{\sin \theta }=\dfrac{{{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta }{\sin \theta }$
Now, we will look at the R.H.S.,
R.H.S. $=\sin \left( {{270}^{\circ }}-\theta \right)=\sin \left( {{180}^{\circ }}+{{90}^{\circ }}-\theta \right)$
Again, we will use the relations mentioned above to simplify the R.H.S. as follows,
$\sin \left( {{180}^{\circ }}+{{90}^{\circ }}-\theta \right)=\sin \left( {{180}^{\circ }}+\left( {{90}^{\circ }}-\theta \right) \right)=-\sin \left( {{90}^{\circ }}-\theta \right)=-\cos \theta $
Now, we will equate the L.H.S. and R.H.S. and get the following equation,
$\dfrac{{{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta }{\sin \theta }=-\cos \theta $
Simplifying the above equation, we get
${{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta =-\cos \theta \sin \theta $
$\therefore x\cdot {{\cos }^{2}}\theta =-{{\sin }^{2}}\theta -\cos \theta \sin \theta $
Dividing by ${{\cos }^{2}}\theta $ on both sides of the above equation, we get
$\begin{align}
& x=-\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-\dfrac{\cos \theta \sin \theta }{{{\cos }^{2}}\theta } \\
& =-{{\tan }^{2}}\theta -\tan \theta
\end{align}$
Hence, the value is $x={{\tan }^{2}}\theta -\tan \theta $.
Note: It is easy to make an error in such types of questions, if we are not aware of the relations that are mentioned in the hint as they are very crucial to solve the problem and get to the solution. In this question, we have used multiple relations and trigonometric identities to simplify the given equation. While solving questions related to trigonometry, simplification of the given equations is the key aspect.
$\begin{align}
& \sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \\
& \sin \left( {{90}^{\circ }}+\theta \right)=\cos \theta \\
& \sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta \\
& \sin \left( {{180}^{\circ }}+\theta \right)=-\sin \theta \\
& \cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \\
& \cos \left( {{90}^{\circ }}+\theta \right)=\sin \theta \\
& \cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta \\
& \cos \left( {{180}^{\circ }}+\theta \right)=-\cos \theta \\
\end{align}$
Now, using these expressions and relations of sine and cosine functions, we will simplify the given equation and find the value of $x$.
Complete step by step answer:
We will first simplify the L.H.S. of the given equation by using the relations mentioned above, in the following manner,
L.H.S. $=\cos \left( 270{}^\circ +\theta \right)+x\cdot \cos \theta \cdot \cot \left( 180{}^\circ +\theta \right)$
Now, we know that $\cot \left( {{180}^{\circ }}+\theta \right)=\dfrac{\cos \left( {{180}^{\circ }}+\theta \right)}{\sin \left( {{180}^{\circ }}+\theta \right)}=\dfrac{-\cos \theta }{-\sin \theta }=\dfrac{\cos \theta }{\sin \theta }$.
The value for $\cos \left( {{270}^{\circ }}+\theta \right)=\cos \left( {{180}^{\circ }}+{{90}^{\circ }}+\theta \right)=\cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)$ . We will use the relations mentioned above in the following manner,
$\cos \left( {{180}^{\circ }}+\left( {{90}^{\circ }}+\theta \right) \right)=-\cos \left( {{90}^{\circ }}+\theta \right)=\sin \theta $. Therefore, we have $\cos \left( {{270}^{\circ }}+\theta \right)=\sin \theta $.
Substituting these values in the L.H.S., we get
L.H.S. $=\sin \theta +x\cdot \cos \theta \cdot \dfrac{\cos \theta }{\sin \theta }$
Simplifying this equation, we get
L.H.S. $=\sin \theta +x\cdot \dfrac{{{\cos }^{2}}\theta }{\sin \theta }=\dfrac{{{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta }{\sin \theta }$
Now, we will look at the R.H.S.,
R.H.S. $=\sin \left( {{270}^{\circ }}-\theta \right)=\sin \left( {{180}^{\circ }}+{{90}^{\circ }}-\theta \right)$
Again, we will use the relations mentioned above to simplify the R.H.S. as follows,
$\sin \left( {{180}^{\circ }}+{{90}^{\circ }}-\theta \right)=\sin \left( {{180}^{\circ }}+\left( {{90}^{\circ }}-\theta \right) \right)=-\sin \left( {{90}^{\circ }}-\theta \right)=-\cos \theta $
Now, we will equate the L.H.S. and R.H.S. and get the following equation,
$\dfrac{{{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta }{\sin \theta }=-\cos \theta $
Simplifying the above equation, we get
${{\sin }^{2}}\theta +x\cdot {{\cos }^{2}}\theta =-\cos \theta \sin \theta $
$\therefore x\cdot {{\cos }^{2}}\theta =-{{\sin }^{2}}\theta -\cos \theta \sin \theta $
Dividing by ${{\cos }^{2}}\theta $ on both sides of the above equation, we get
$\begin{align}
& x=-\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-\dfrac{\cos \theta \sin \theta }{{{\cos }^{2}}\theta } \\
& =-{{\tan }^{2}}\theta -\tan \theta
\end{align}$
Hence, the value is $x={{\tan }^{2}}\theta -\tan \theta $.
Note: It is easy to make an error in such types of questions, if we are not aware of the relations that are mentioned in the hint as they are very crucial to solve the problem and get to the solution. In this question, we have used multiple relations and trigonometric identities to simplify the given equation. While solving questions related to trigonometry, simplification of the given equations is the key aspect.
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE

The area of a 6m wide road outside a garden in all class 10 maths CBSE

What is the electric flux through a cube of side 1 class 10 physics CBSE

If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE

The radius and height of a cylinder are in the ratio class 10 maths CBSE

An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE

Trending doubts
Who was Subhash Chandra Bose Why was he called Net class 10 english CBSE

Write an application to the principal requesting five class 10 english CBSE

What are the public facilities provided by the government? Also explain each facility

What is Commercial Farming ? What are its types ? Explain them with Examples

Complete the sentence with the most appropriate word class 10 english CBSE

Find the area of the minor segment of a circle of radius class 10 maths CBSE
