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 10743 by okhema last updated on 24/Feb/17

$$showthat\sec {}^2\theta =\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta -\sin }$$

Commented by ridwan balatif last updated on 24/Feb/17

$$\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta -\sin \theta }=\frac{\frac{1}{\sin \theta }}{\frac{1}{\sin \theta }-\sin \theta }$$$$=\frac{\frac{1}{\sin \theta }}{\frac{1-{\sin }^2\theta }{\sin \theta }}$$$$=\frac{\frac{\frac{1}{\sin \theta }}{{\cos }^2\theta }}{\sin \theta }$$$$=\frac{1}{\sin \theta }\times \frac{\sin \theta }{{\cos }^2\theta }$$$$=\frac{1}{{\cos }^2\theta }$$$$\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta -\sin \theta }={\sec }^2\theta$$$$\mathrm{remember}!$$$$\sec \theta =\frac{1}{\cos \theta }$$$$\mathrm{cosec}\theta =\frac{1}{\sin \theta }$$

Commented by okhema last updated on 24/Feb/17

$$yesthatswhatigotbutwasntsureifitscorrect$$$$causeionlyknow{sec}^2\theta =1+{tan}^2\theta butthanksanyways$$

Answered by ridwan balatif last updated on 24/Feb/17

$${\sec }^2\theta =\frac{1}{{\cos }^2\theta }$$$$=\left(\frac{1}{1-{\sin }^2\theta }\right)\left(\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta }\right)$$$$=\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta -\mathrm{cosec}\theta \times \sin \theta \times \sin \theta }$$$$=\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta -\frac{1}{\sin \theta }\times \sin \theta \times \sin \theta }$$$${\sec }^2\theta =\frac{\mathrm{cosec}\theta }{\mathrm{cosec}\theta -\sin \theta }$$$$\mathrm{PROVED}!!!!$$$$$$

Commented by okhema last updated on 24/Feb/17

$$whycouldntyoustartfromtheright...causethatswhatidobydoingtherighthandsidebutiendupwith\frac{1}{{cos}^2\theta }$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com