Question

Prove that $(cosec\theta -sin\theta )(sec\theta -cos\theta )=\frac{1}{tan\theta +cot\theta }$

Answer 1

Determine the proving of the expression $(cosec\theta -sin\theta )(sec\theta -cos\theta )=\frac{1}{tan\theta +cot\theta }$

Solve the L.H.S part:

$$\begin{gathered} (cosec\theta -sin\theta )(sec\theta -cos\theta )=(\frac{1}{sin\theta }-sin\theta )(\frac{1}{\cos \theta }-cos\theta )\left[\because \mathrm{cosec}\theta =\frac{1}{sin\theta }and\sec \theta =\frac{1}{cos\theta }\right]\mathbf{} \\ \Rightarrow =\left(\frac{1-si{n}^2\theta }{sin\theta }\right)\left(\frac{1-co{s}^2\theta }{\cos \theta }\right)\left[\because si{n}^2\theta +co{s}^2\theta =1\right] \\ \Rightarrow =\frac{co{s}^2\theta }{sin\theta }\times \frac{{\sin }^2\theta }{\cos \theta } \\ \Rightarrow =cos\theta sin\theta \end{gathered}$$

Solve the R.H.S part:

$$\begin{gathered} \frac{1}{tan\theta +cot\theta }=\frac{1}{{\displaystyle \frac{\sin \theta }{\cos \theta }}+{\displaystyle \frac{\cos \theta }{\sin \theta }}}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{tan}\mathbf{}\mathbf{\theta }\mathbf{=}\mathbf{}\frac{\mathbf{sin}\mathbf{}\mathbf{\theta }}{\mathbf{cos}\mathbf{}\mathbf{\theta }\mathbf{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{}\mathbf{}\mathbf{cot}\mathbf{}\mathbf{\theta }\mathbf{=}\mathbf{}\frac{\mathbf{cos}\mathbf{}\mathbf{\theta }}{\mathbf{sin}\mathbf{}\mathbf{\theta }}\mathbf{}\mathbf{]} \\ \Rightarrow =\frac{1}{{\displaystyle \frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }}} \\ \Rightarrow =\frac{\cos \theta \sin \theta }{{\displaystyle {\sin }^2\theta +{\cos }^2\theta }}\mathbf{[}\mathbf{}\mathbf{\because }\mathbf{}{\mathbf{sin}}^{\mathbf{2}}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{}{\mathbf{cos}}^{\mathbf{2}}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{]} \\ \Rightarrow =\frac{\cos \theta \sin \theta }{{\displaystyle 1}} \\ \Rightarrow =\cos \theta \sin \theta \end{gathered}$$

Hence,it is proved that $(cosec\theta -sin\theta )(sec\theta -cos\theta )=\frac{1}{tan\theta +cot\theta }$