Question

Prove the following identity: $\frac{\cot \theta }{cosec\theta +1}+\frac{cosec\theta +1}{\cot \theta }=2\sec \theta$

Answer 1

Prove the following identity.

Given data: We have to prove that, $\frac{\cot \theta }{cosec\theta +1}+\frac{cosec\theta +1}{\cot \theta }=2\sec \theta$

Using the Trigonometric identities:

$$\begin{gathered} 1+\mathrm{co}{t}^2\theta =\cos e{c}^2\theta \_\_\_\_\_\_\left(1\right) \\ \cos ec\theta =\frac{1}{\sin \theta }\_\_\_\_\_\_\left(2\right) \\ \frac{1}{\cos \theta }=sec\theta \_\_\_\_\_\_\_\left(3\right) \end{gathered}$$

Proof of the given expression:

Evaluating LHS,

$$\begin{gathered} \frac{\cot \theta }{cosec\theta +1}+\frac{cosec\theta +1}{\cot \theta } \\ =\frac{\cot \theta \times \cot \theta +(cosec\theta +1)(cosec\theta +1)}{\cot \theta (cosec\theta +1)} \\ =\frac{{\cot }^2\theta +\mathrm{cose}{c}^2\theta +1+2cosec\theta }{{\displaystyle \frac{\cos \theta }{\sin \theta }}\left(\frac{1}{\sin \theta }+1\right)} \\ =\frac{{\cot }^2\theta +1+\mathrm{cose}{c}^2\theta +2cosec\theta }{{\displaystyle \frac{\cos \theta }{\sin \theta }\left(\frac{1}{\sin \theta }+1\right)}} \\ =\frac{\mathrm{cose}{c}^2\theta +\mathrm{cose}{c}^2\theta +2cosec\theta }{{\displaystyle \frac{\cos \theta }{\sin \theta }\left(\frac{1}{\sin \theta }+1\right)}} \\ =\frac{2\mathrm{cose}{c}^2\theta +2cosec\theta }{{\displaystyle \frac{\cos \theta }{\sin \theta }\left(\frac{1}{\sin \theta }+1\right)}} \\ =\frac{2\mathrm{cose}c\theta (cosec\theta +1)}{{\displaystyle \frac{\cos \theta }{\sin \theta }\left(\frac{1}{\sin \theta }+1\right)}} \\ =\frac{{\displaystyle \frac{2}{\sin \theta }\left(\frac{1}{\sin \theta }+1\right)}}{{\displaystyle \frac{\cos \theta }{\sin \theta }\left(\frac{1}{\sin \theta }+1\right)}} \\ =\frac{2}{\cos \theta } \\ =2sec\theta =RHS \end{gathered}$$

Since, LHS = RHS

Hence, the given expression is proved.