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Question: If \(1+\cot \theta =\cos ec\theta \), then the general value of \(\theta \) is \(1)\text{ }n\pi +\...

If 1+cotθ=cosecθ1+\cot \theta =\cos ec\theta , then the general value of θ\theta is
1) nπ+π21)\text{ }n\pi +\dfrac{\pi }{2}
3) 2nππ23)\text{ 2}n\pi -\dfrac{\pi }{2}
3) 2nπ+π23)\text{ 2}n\pi +\dfrac{\pi }{2}
4) None of these4)\text{ None of these}

Explanation

Solution

In this question we have been given with a trigonometric expression for which we have to find the general value of θ\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 sinθ\sin \theta and cosθ\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:
1+cotθ=cosecθ\Rightarrow 1+\cot \theta =\cos ec\theta
Now we know that cotθ=cosθsinθ\cot \theta =\dfrac{\cos \theta }{\sin \theta } and cosecθ=1sinθ\cos ec\theta =\dfrac{1}{\sin \theta } therefore, on substituting, we get:
1+cosθsinθ=1sinθ\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:
sinθ+cosθsinθ=1sinθ\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:
sinθ+cosθ=1\Rightarrow \sin \theta +\cos \theta =1
On squaring both the sides, we get:
(sinθ+cosθ)2=12\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}={{1}^{2}}
On expanding the terms, we get:\
sin2θ+2sinθcosθ+cos2θ=12\Rightarrow {{\sin }^{2}}\theta +2\sin \theta \cos \theta +{{\cos }^{2}}\theta ={{1}^{2}}
Now we know the identity that sin2θ+cos2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 therefore, we can write:
1+2sinθcosθ=1\Rightarrow 1+2\sin \theta \cos \theta =1
Now we know the formula sin2θ=2sinθcosθ\sin 2\theta =2\sin \theta \cos \theta therefore, we get:
sin2θ=0\Rightarrow \sin 2\theta =0
Now we know the general solution that when sin2θ=0\sin 2\theta =0, we have the value of θ\theta as:
θ=2nπ+π2\Rightarrow \theta =2n\pi +\dfrac{\pi }{2}, which is the required value.
Therefore, the correct option is (3)\left( 3 \right).

Note: To simplify any given equation, it is good practice to convert all the identities into sinθ\sin \theta and cosθ\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.