Please enter the following details
Free Daily Targets & Mocks
Study Material & Formulas PDFs
1 on 1 Mentorship
Solution
Given that,
$$\cos \theta = \frac{2p}{1 + p^2}$$
$$\sin \theta=\sqrt{1-\cos^2\theta}=\sqrt{1-(\frac{2p}{1 + p^2})^2}=\sqrt{\dfrac{(1+p^2)^2-4p^2}{(1+p^2)^2}}$$
$$\sin \theta=\sqrt{\dfrac{1+p^4+2p^2-4p^2}{(1+p^2)^2}}=\sqrt{\dfrac{1+p^4-2p^2}{(1+p^2)^2}}=\sqrt{\dfrac{(1-p^2)^2}{(1+p^2)^2}}=\dfrac{1-p^2}{1+p^2}$$
So, $$\tan \theta=\dfrac{\dfrac{1-p^2}{1+p^2}}{ \frac{2p}{1 + p^2}}=\dfrac{1-p^2}{2p}$$
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
