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 82767 by jagoll last updated on 24/Feb/20

$$\mathrm{If}\frac{\sin (\mathrm{A}+\theta )}{\sin (B+\theta )}=\sqrt{\frac{\sin 2A}{\sin 2B}}$$$$provethat\tan {}^2\theta =\tan A.\tan B$$

Commented by jagoll last updated on 24/Feb/20

$$thankyoumrwandjohn$$

Answered by mr W last updated on 24/Feb/20

$$\frac{\sin A\cos \theta +\cos A\sin \theta }{\sin B\cos \theta +\cos B\sin \theta }=\sqrt{\frac{\sin A\cos A}{\sin B\cos B}}$$$$\frac{1+\frac{\tan \theta }{\tan A}}{1+\frac{\tan \theta }{\tan B}}=\frac{\sin B}{\sin A}\sqrt{\frac{\sin A\cos A}{\sin B\cos B}}$$$$\frac{1+\frac{\tan \theta }{\tan A}}{1+\frac{\tan \theta }{\tan B}}=\sqrt{\frac{\tan B}{\tan A}}$$$$1+\frac{\tan \theta }{\tan A}=\sqrt{\frac{\tan B}{\tan A}}+\frac{\tan \theta }{\sqrt{\tan A\tan B}}$$$$(\frac{1}{\tan A}-\frac{1}{\sqrt{\tan A\tan B}})\tan \theta =\sqrt{\frac{\tan B}{\tan A}}-1$$$$(\frac{1}{\sqrt{\tan A}}-\frac{1}{\sqrt{\tan B}})\tan \theta =\sqrt{\tan B}-\sqrt{\tan A}$$$$\frac{\tan \theta }{\sqrt{\tan A\tan B}}=1$$$$\tan \theta =\sqrt{\tan A\tan B}$$$$\Rightarrow {\tan }^2\theta =\tan A\tan B$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com