Answer
Verified
396.6k+ views
Hint: We solve this problem by converting either sin to cos (or) cos to sin using the formula $\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)$ and then we equate both the angles. After changing the angles into the same trigonometric identity, we equate the angles and we solve them to find the value of $\theta $.
Complete step by step answer:
We use the formula,
$\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)$
Let us consider the given trigonometric equation
$\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\cos \left( \dfrac{\pi }{4}.\tan \theta \right)$
Now, we convert cos into sin to equate the angles. So, we get,
$\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{4}.\tan \theta \right)$
As the trigonometric terms on both sides are equal. By equating both the angles, we get
$\begin{align}
& \Rightarrow \dfrac{\pi }{4}.\cot \theta =\dfrac{\pi }{2}-\dfrac{\pi }{4}.\tan \theta \\
& \Rightarrow \dfrac{\pi }{4}.\cot \theta +\dfrac{\pi }{4}.\tan \theta =\dfrac{\pi }{2} \\
& \Rightarrow \dfrac{\pi }{4}\left( \cot \theta +\tan \theta \right)=\dfrac{\pi }{2} \\
\end{align}$
Now we convert $\cot \theta $ to $\tan \theta $ , we get
$\dfrac{\pi }{4}\left( \dfrac{1}{\tan \theta }+\tan \theta \right)=\dfrac{\pi }{2}$
After cancelling $\dfrac{\pi }{4}$ on both the sides, we get
$\begin{align}
& \Rightarrow \dfrac{1}{\tan \theta }+\tan \theta =2 \\
& \Rightarrow \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2 \\
& \Rightarrow 1+{{\tan }^{2}}\theta =2\tan \theta \\
& \Rightarrow {{\tan }^{2}}\theta -2\tan \theta +1=0 \\
\end{align}$
Using the formula of ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we can factorize the above quadratic equation as,
$\begin{align}
& \Rightarrow {{\left( \tan \theta -1 \right)}^{2}}=0 \\
& \Rightarrow \tan \theta -1=0 \\
& \Rightarrow \tan \theta =1 \\
& \Rightarrow \tan \theta =\tan \dfrac{\pi }{4} \\
\end{align}$
But we have to find the general solution for $\tan \theta =1$.
Let us consider the formula for finding the general solution.
If $\alpha $ is the principal solution of $\tan \theta =k$, then the general solution of $\tan \theta $ is $n\pi +\alpha $.
Using the above formula we can write the general solution for $\tan \theta =1$ as,
The general solution for $\tan \theta =\tan \dfrac{\pi }{4}$ is $n\pi +\left( \dfrac{\pi }{4} \right)$
Therefore, if $\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\cos \left( \dfrac{\pi }{4}.\tan \theta \right)$, then $\theta =n\pi +\left( \dfrac{\pi }{4} \right)$
So, the correct answer is “Option C”.
Note: One can make a mistake by considering the general of solution of $\tan \theta =1$ as $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ instead of $n\pi +\left( \dfrac{\pi }{4} \right)$. So, one need to be careful while writing the general terms for trigonometric identities. For example, if we consider $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ as the general solution for $\tan \theta =1$, then by taking n=1, $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ gives us -1, which is not the solution for $\tan \theta =1$. So, one must be careful while writing the general solution for the trigonometric identities.
Complete step by step answer:
We use the formula,
$\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)$
Let us consider the given trigonometric equation
$\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\cos \left( \dfrac{\pi }{4}.\tan \theta \right)$
Now, we convert cos into sin to equate the angles. So, we get,
$\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\dfrac{\pi }{4}.\tan \theta \right)$
As the trigonometric terms on both sides are equal. By equating both the angles, we get
$\begin{align}
& \Rightarrow \dfrac{\pi }{4}.\cot \theta =\dfrac{\pi }{2}-\dfrac{\pi }{4}.\tan \theta \\
& \Rightarrow \dfrac{\pi }{4}.\cot \theta +\dfrac{\pi }{4}.\tan \theta =\dfrac{\pi }{2} \\
& \Rightarrow \dfrac{\pi }{4}\left( \cot \theta +\tan \theta \right)=\dfrac{\pi }{2} \\
\end{align}$
Now we convert $\cot \theta $ to $\tan \theta $ , we get
$\dfrac{\pi }{4}\left( \dfrac{1}{\tan \theta }+\tan \theta \right)=\dfrac{\pi }{2}$
After cancelling $\dfrac{\pi }{4}$ on both the sides, we get
$\begin{align}
& \Rightarrow \dfrac{1}{\tan \theta }+\tan \theta =2 \\
& \Rightarrow \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2 \\
& \Rightarrow 1+{{\tan }^{2}}\theta =2\tan \theta \\
& \Rightarrow {{\tan }^{2}}\theta -2\tan \theta +1=0 \\
\end{align}$
Using the formula of ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, we can factorize the above quadratic equation as,
$\begin{align}
& \Rightarrow {{\left( \tan \theta -1 \right)}^{2}}=0 \\
& \Rightarrow \tan \theta -1=0 \\
& \Rightarrow \tan \theta =1 \\
& \Rightarrow \tan \theta =\tan \dfrac{\pi }{4} \\
\end{align}$
But we have to find the general solution for $\tan \theta =1$.
Let us consider the formula for finding the general solution.
If $\alpha $ is the principal solution of $\tan \theta =k$, then the general solution of $\tan \theta $ is $n\pi +\alpha $.
Using the above formula we can write the general solution for $\tan \theta =1$ as,
The general solution for $\tan \theta =\tan \dfrac{\pi }{4}$ is $n\pi +\left( \dfrac{\pi }{4} \right)$
Therefore, if $\sin \left( \dfrac{\pi }{4}.\cot \theta \right)=\cos \left( \dfrac{\pi }{4}.\tan \theta \right)$, then $\theta =n\pi +\left( \dfrac{\pi }{4} \right)$
So, the correct answer is “Option C”.
Note: One can make a mistake by considering the general of solution of $\tan \theta =1$ as $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ instead of $n\pi +\left( \dfrac{\pi }{4} \right)$. So, one need to be careful while writing the general terms for trigonometric identities. For example, if we consider $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ as the general solution for $\tan \theta =1$, then by taking n=1, $n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4}$ gives us -1, which is not the solution for $\tan \theta =1$. So, one must be careful while writing the general solution for the trigonometric identities.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE