Category: Trigonometry

Question Number 63060 by rajesh4661kumar@gamil.com last updated on 28/Jun/19 Answered by Hope last updated on 28/Jun/19 $$\sqrt{\frac{{(1+cos\theta )}^2}{{sin}^2\theta }}$$$$=\mid \frac{1+cos\theta }{sin\theta }\mid$$$$=\frac{\mid 1+cos\theta \mid }{\mid sin\theta \mid }$$$${when}\:\:\:\pi>\theta>\mathrm{0}\:\:{so}\:{sin}\theta=+{ve}…

Question Number 128546 by mr W last updated on 08/Jan/21 $$findasimpleexpressionfor$$$$y={\sin }^{-1}\left(\sin x\right)$$$$x\in \mathbb{R}andxinrad.$$ Commented by liberty last updated on 08/Jan/21…

Question Number 128336 by liberty last updated on 06/Jan/21 $$\mathrm{If}\frac{\sec \mathrm{x}+\tan \mathrm{x}}{\sec \mathrm{x}-\tan \mathrm{x}}=5\mathrm{then}\mathrm{what}\mathrm{is}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\frac{3\cos 2\mathrm{x}+1}{3\cos 2\mathrm{x}-1}?$$ Answered by bramlexs22 last updated on 06/Jan/21 $$\Rightarrow \frac{\sec x+\tan x}{\sec x-\tan x}=5$$$$\Rightarrow\frac{\mathrm{cos}\:{x}\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)}{\mathrm{cos}\:{x}\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)}\:=\:\mathrm{5} \…

Question Number 62761 by lalitchand last updated on 25/Jun/19 $$\mathrm{if}\cos \theta =\frac{\mathrm{cosA}.\mathrm{cosB}}{1-\mathrm{cosA}.\mathrm{cosB}}\mathrm{prove}\mathrm{that}\mathrm{Tan}\frac{\theta }{2}=\mathrm{Tan}\frac{\mathrm{A}}{2}.\mathrm{Cot}\frac{\mathrm{B}}{2}$$ Commented by Prithwish sen last updated on 25/Jun/19 $$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{is}$$$$\cos \theta =\frac{\mathrm{cosA}-\mathrm{cosB}}{1-\mathrm{cosAcosB}}$$$$\frac{\mathrm{1}−\mathrm{cos}\theta}{\mathrm{1}+\mathrm{cos}\theta}\:=\:\frac{\mathrm{1}−\mathrm{cosA}+\mathrm{cosB}−\mathrm{cosAcosB}}{\mathrm{1}+\mathrm{cosA}−\mathrm{cosB}−\mathrm{cosAcosB}}…

Question Number 128178 by bemath last updated on 05/Jan/21 $$\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{any}$$$$\mathrm{quadrilateral}\mathrm{ABCD}\mathrm{is}$$$$\sqrt{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})(\mathrm{s}-\mathrm{d})-\mathrm{abcd}\cos {}^2\left(\frac{\mathrm{A}+\mathrm{C}}{2}\right)}.$$ Answered by mr W last updated on 05/Jan/21…