This is beta site. https://beta.mathcity.org/ 2026-06-07T06:07:00+00:00 https://beta.mathcity.org/ https://beta.mathcity.org/_media/logo.png text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p2?rev=1737476040&do=diff Question 2, Exercise 8.1 Solutions of Question 2 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\cos 15^{\circ}$$\cos \left(45^{\circ}-30^{\circ}\right)$\begin{align*} \cos 15^{\circ} & = \cos \left(45^{\circ}-30^{\circ}\right)\\ &= \cos 45 \cos 30 + \sin 45 \sin 30 \\ &= \dfrac{1}{\sqrt{2}}\cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p3?rev=1737476040&do=diff Question 3, Exercise 8.1 Solutions of Question 3 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\cos 120^{\circ}$$\cos \left(180^{\circ}-60^{\circ}\right)$$\cos \left(90^{\circ}+30^{\circ}\right)$\begin{align*} \cos 120^{\circ} & = \cos \left(180^{\circ}-60^{\circ}\right) \\ &= - \cos 60 ^{\circ}\\ &= -\dfrac{1}{2}. \end{align*}\begin{… text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/ex8-1-p13?rev=1737476040&do=diff Question 14, Exercise 8.1 Solutions of Question 14 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\theta$$\sin \theta$$\cos \theta$$\alpha$\begin{align*} &\tan\alpha = \frac{\overline{BC}}{\overline{AB}} \\ \implies &\tan\alpha = \frac{3}{3} = 1 \\ \implies &\alpha = \tan^{-1}(1) = 45^\circ \end{align*}$45^\circ$$\theta$\begin{align*} … text/html 2025-01-21T16:14:00+00:00 Anonymous (anonymous@undisclosed.example.com) https://beta.mathcity.org/math-11-nbf/sol/unit08/re-ex-p1?rev=1737476040&do=diff Question 1, Review Exercise Solutions of Question 1 of Review Exercise of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. $\sin \left(45^{\circ}-30^{\circ}\right)=\ldots$$\frac{\sqrt{6}-\sqrt{2}}{4}$$\frac{\sqrt{6}+\sqrt{2}}{4}$$\frac{\sqrt{6}-\sqrt{2}}{2}$$\frac{\sqrt{3}-\sqrt{2}}{2}$$\tan \left(\frac{\pi}{6}+\frac{\pi}{4}\right)=\ldots$$\frac{\sqrt{3}-1}…