Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 31679 by pieroo last updated on 12/Mar/18

$$\mathrm{prove}\mathrm{that}{\sec }^2\theta {\mathrm{cosec}}^2\theta ={\sec }^2\theta +{\mathrm{cosec}}^2\theta$$

Answered by Tinkutara last updated on 12/Mar/18

$${\sec }^2\theta {\mathrm{cosec}}^2\theta =\frac{1}{{\sin }^2\theta {\cos }^2\theta }$$$$=\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }$$$$=\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }={\sec }^2\theta +{\mathrm{cosec}}^2\theta$$

Commented by pieroo last updated on 12/Mar/18

$$\mathrm{thanks}\mathrm{boss}$$

Answered by Joel578 last updated on 12/Mar/18

$${\sec }^2\theta {\mathrm{cosec}}^2\theta =(1+{\tan }^2\theta )(1+{\cot }^2\theta )$$$$=1+{\cot }^2\theta +{\tan }^2\theta +{\tan }^2\theta {\cot }^2\theta$$$$={\mathrm{cosec}}^2\theta +{\sec }^2\theta$$

Commented by pieroo last updated on 12/Mar/18

$$\mathrm{i}\mathrm{am}\mathrm{grateful}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com