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Question: Find the value of \(1+{{\cot }^{2}}\theta \) A. \({{\sec }^{2}}\theta \) B. \({{\operatorname{c...

Find the value of 1+cot2θ1+{{\cot }^{2}}\theta
A. sec2θ{{\sec }^{2}}\theta
B. cosec2θ{{\operatorname{cosec}}^{2}}\theta
C. tan2θ{{\tan }^{2}}\theta
D. 1

Explanation

Solution

To solve this question we will use the trigonometric identities and trigonometric formulas. Following trigonometric rules will be used to solve this question:
cotθ=cosθsinθ sin2θ+cos2θ=1 1sinθ=cosecθ \begin{aligned} & \cot \theta =\dfrac{\cos \theta }{\sin \theta } \\\ & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\\ & \dfrac{1}{\sin \theta }=cosec\theta \\\ \end{aligned}

Complete answer:
We have to solve 1+cot2θ1+{{\cot }^{2}}\theta .
We know that cotθ=cosθsinθ\cot \theta =\dfrac{\cos \theta }{\sin \theta }
Now, substituting the value in the given equation we get

& \Rightarrow 1+{{\cot }^{2}}\theta \\\ & \Rightarrow 1+{{\left( \dfrac{\cos \theta }{\sin \theta } \right)}^{2}} \\\ \end{aligned}$$ Now, solving further we get $$\Rightarrow 1+\left( \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta } \right)$$ By taking LCM and solving further we get $$\Rightarrow \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }$$ Now, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Now, substituting the value in the above equation we get $$\Rightarrow \dfrac{1}{{{\sin }^{2}}\theta }$$ Now, we know that $\dfrac{1}{\sin \theta }=cosec\theta $ Substituting the value in the above equation we get $\Rightarrow cose{{c}^{2}}\theta $ So, we get the value of $1+{{\cot }^{2}}\theta =cose{{c}^{2}}\theta $. **Option B is the correct answer.** **Note:** To solve such types of questions always remember the trigonometric functions and identities. Also be careful while conversion of functions, please avoid mistakes. The key concept to solve this question is to simplify the given equation using trigonometric functions.