Question

17) Prove the following identity :

(cosec $\theta$ - sin $\theta$). (sec $\theta$ - cos $\theta$). (tan $\theta$ + cot $\theta$) = 1

Answer 1

$$\begin{gathered} \mathrm{LHS}=\left(\mathrm{cosec}\mathrm{\theta }-\sin \mathrm{\theta }\right)\left(\sec \mathrm{\theta }-\cos \mathrm{\theta }\right)\left(\tan \mathrm{\theta }+\cot \mathrm{\theta }\right) \\ =\left(\frac{1}{\sin \mathrm{\theta }}-\sin \mathrm{\theta }\right)\left(\frac{1}{\cos \mathrm{\theta }}-\cos \mathrm{\theta }\right)\left(\frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}+\frac{\cos \mathrm{\theta }}{\sin \mathrm{\theta }}\right) \\ =\left(\frac{1-{\sin }^2\mathrm{\theta }}{\sin \mathrm{\theta }}\right)\left(\frac{1-{\cos }^2\mathrm{\theta }}{\cos \mathrm{\theta }}\right)\left(\frac{{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }.\cos \mathrm{\theta }}\right) \\ =\frac{{\cos }^2\mathrm{\theta }}{\sin \mathrm{\theta }}\times \frac{{\sin }^2\mathrm{\theta }}{\cos \mathrm{\theta }}\times \frac{1}{\sin \mathrm{\theta }.\cos \mathrm{\theta }} \\ =\frac{{\sin }^2\mathrm{\theta }.{\cos }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }.{\cos }^2\mathrm{\theta }} \\ =1 \\ =\mathrm{RHS} \end{gathered}$$