
If $1+\cot \theta =\cos ec\theta $, then the general value of $\theta $ is
$1)\text{ }n\pi +\dfrac{\pi }{2}$
$3)\text{ 2}n\pi -\dfrac{\pi }{2}$
$3)\text{ 2}n\pi +\dfrac{\pi }{2}$
$4)\text{ None of these}$
Answer
488.1k+ views
Hint: In this question we have been given with a trigonometric expression for which we have to find the general value of $\theta $. We will solve this question by simplifying the terms in the expression and then rearranging the terms to get the expression in the form of $\sin \theta $ and $\cos \theta $. We will then use the double angle formula to further simplify the expression and get the required solution.
Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow 1+\cot \theta =\cos ec\theta $
Now we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\cos ec\theta =\dfrac{1}{\sin \theta }$ therefore, on substituting, we get:
$\Rightarrow 1+\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\sin \theta }$
On taking the lowest common multiple on the left-hand side of the expression, we get:
$\Rightarrow \dfrac{\sin \theta +\cos \theta }{\sin \theta }=\dfrac{1}{\sin \theta }$
Now since the denominator on both the sides is same, we cancel them and write it as:
$\Rightarrow \sin \theta +\cos \theta =1$
On squaring both the sides, we get:
$\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}={{1}^{2}}$
On expanding the terms, we get:\
$\Rightarrow {{\sin }^{2}}\theta +2\sin \theta \cos \theta +{{\cos }^{2}}\theta ={{1}^{2}}$
Now we know the identity that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ therefore, we can write:
$\Rightarrow 1+2\sin \theta \cos \theta =1$
Now we know the formula $\sin 2\theta =2\sin \theta \cos \theta $ therefore, we get:
$\Rightarrow \sin 2\theta =0$
Now we know the general solution that when $\sin 2\theta =0$, we have the value of $\theta $ as:
$\Rightarrow \theta =2n\pi +\dfrac{\pi }{2}$, which is the required value.
Therefore, the correct option is $\left( 3 \right)$.
Note: To simplify any given equation, it is good practice to convert all the identities into $\sin \theta $ and $\cos \theta $ for simplifying. If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation. The various trigonometric identities and formulae should be remembered while doing these types of sums.
Complete step-by-step solution:
We have the expression given to us as:
$\Rightarrow 1+\cot \theta =\cos ec\theta $
Now we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\cos ec\theta =\dfrac{1}{\sin \theta }$ therefore, on substituting, we get:
$\Rightarrow 1+\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\sin \theta }$
On taking the lowest common multiple on the left-hand side of the expression, we get:
$\Rightarrow \dfrac{\sin \theta +\cos \theta }{\sin \theta }=\dfrac{1}{\sin \theta }$
Now since the denominator on both the sides is same, we cancel them and write it as:
$\Rightarrow \sin \theta +\cos \theta =1$
On squaring both the sides, we get:
$\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}={{1}^{2}}$
On expanding the terms, we get:\
$\Rightarrow {{\sin }^{2}}\theta +2\sin \theta \cos \theta +{{\cos }^{2}}\theta ={{1}^{2}}$
Now we know the identity that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ therefore, we can write:
$\Rightarrow 1+2\sin \theta \cos \theta =1$
Now we know the formula $\sin 2\theta =2\sin \theta \cos \theta $ therefore, we get:
$\Rightarrow \sin 2\theta =0$
Now we know the general solution that when $\sin 2\theta =0$, we have the value of $\theta $ as:
$\Rightarrow \theta =2n\pi +\dfrac{\pi }{2}$, which is the required value.
Therefore, the correct option is $\left( 3 \right)$.
Note: To simplify any given equation, it is good practice to convert all the identities into $\sin \theta $ and $\cos \theta $ for simplifying. If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation. The various trigonometric identities and formulae should be remembered while doing these types of sums.
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