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Question 10, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:42 +0000

Question 10, Exercise 8.1 Solutions of Question 10 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. $\sin \left(\dfrac{\pi}{2}-\alpha\right)=\cos \alpha$\begin{align*} L.H.S & = \sin \left(\frac{\pi}{2}-\alpha\right) \\ & =\sin\frac{\pi}{2} \cos \alpha - \cos \frac{\pi}{2} \sin\alpha \\ & = 1\times \cos \alpha - 0 \times \sin\alpha \\ & =…

Question 11, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:31 +0000

Question 11, Exercise 8.1 Solutions of Question 11 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. $\dfrac{\sin \left(180^{\circ}+\lambda\right) \cos \left(270^{\circ}+\lambda\right)}{\sin \left(180^{\circ}-\lambda\right) \cos \left(270^{\circ}-\lambda\right)}=1$\begin{align*} L.H.S & = \dfrac{\sin \left(180^{\circ}+\lambda\right) \cos \…

Question 1, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:30 +0000

Question 1, Exercise 8.1 Solutions of Question 1 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 (\alpha \pm \beta), \sin (\alpha \pm \beta)$$\tan (\alpha \pm \beta)$$\alpha=180^{\circ}, \beta=60^{\circ}$$\alpha=180^{\circ}$$\beta=60^{\circ}$\begin{align*} \cos (\alpha + \beta) & = \cos \alpha \cos \beta - \sin \alpha \sin \beta \…

Question 3, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:37 +0000

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{…

Question 4, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:38 +0000

Question 4, Exercise 8.1 Solutions of Question 4 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. $\cos 6 \theta \cos 3 \theta-\sin 6 \theta \sin 3 \theta$\begin{align*} & \cos 6 \theta \cos 3 \theta-\sin 6 \theta \sin 3 \theta \\ & = \cos (6\theta +3\theta) \\ & = \cos 9\theta . \end{align*}$\cos 7 \theta \cos 2 \theta+\sin 7 \theta \sin…

Question 4 Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:40 +0000

Question 4 Exercise 8.2 Solutions of Question 4 of Exercise 8.2 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 2 \theta$$\cos 2 \theta$$\tan 2 \theta$$\sin \frac{\theta}{2}$$\cos \frac{\theta}{2}$$\tan \frac{\theta}{2}$$\cos \theta=\frac{3}{5}$$0<\theta<\frac{\pi}{2}$$\cos\theta=\dfrac{3}{5}$$0<\theta<\dfrac{\pi}{2}$$\theta$$$\sin\theta = \pm \sq…

Question 6 Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:45 +0000

Question 6 Exercise 8.2 Solutions of Question 6 of Exercise 8.2 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 15^{\circ} \cos 15^{\circ}$$$\sin 2 \theta = 2\sin\theta \cos\theta$$$$\sin\theta \cos\theta = \frac{1}{2}\sin 2\theta$$$\theta = 15^{\circ}$\begin{align*} \sin 15^{\circ} \cos 15^{\circ} & = \frac{1}{2}\sin 2(15^{\circ}) \\ & \frac{1}{2…

Question 2(i, ii, iii, iv and v) Exercise 8.3

anonymous@undisclosed.example.com (Anonymous) — Mon, 28 Oct 2024 17:14:31 +0000

Question 2(i, ii, iii, iv and v) Exercise 8.3 Solutions of Question 2(i, ii, iii, iv and v) of Exercise 8.3 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. $\sin 70^{\circ} + \sin 30^{\circ}$\begin{align*} & \quad \sin 70^{\circ} + \sin 30^{\circ} \\ & = 2 \sin \left(\frac{70+30}{2} \right) \cos \left(\frac{70-30}{2} \right) \\ & = 2 \sin \left(\frac{1…

Question 3(i, ii, iii, iv & v) Exercise 8.3

anonymous@undisclosed.example.com (Anonymous) — Mon, 28 Oct 2024 17:14:31 +0000

Question 3(i, ii, iii, iv & v) Exercise 8.3 Solutions of Question 3(i, ii, iii, iv & v) of Exercise 8.3 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. $\dfrac{\cos (\alpha + \beta)}{\cos(\alpha - \beta)}=\dfrac{1- \tan \alpha \tan \beta}{1+ \tan \alpha \tan \beta}$\begin{align*} RHS & = \dfrac{1- \tan \alpha \tan \beta}{1+ \tan \alpha \tan \beta} \\ & …

Question 3(vi, vii, viii, ix & x) Exercise 8.3

anonymous@undisclosed.example.com (Anonymous) — Mon, 28 Oct 2024 17:14:32 +0000

Question 3(vi, vii, viii, ix & x) Exercise 8.3 Solutions of Question 3(vi, vii, viii, ix & x) of Exercise 8.3 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. $2\tan y \cos 3y= \sec y(\sin 4y-\sin 2y)$\begin{align*} LHS & = 2\tan y \cos 3y \\ & = 2 \cdot \frac{\sin y}{\cos y} \cos 3y \\ & = \sec y (2 \cos 3y \sin y) \\ & = \sec y \left(\sin (3y+y)-\sin (…

Question 2, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:35 +0000

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…

Question 5 Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:42 +0000

Question 5 Exercise 8.2 Solutions of Question 5 of Exercise 8.2 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 \theta$$\cos \theta$$\tan \theta$$\sin 2 \theta=\frac{24}{25}, 2 \theta$$\sin 2\theta=\dfrac{24}{25}$$2\theta$$$\cos 2\theta = \pm \sqrt{1-\sin^2 2\theta}$$$2\theta$$\cos 2\theta$\begin{align*}\cos 2\theta & = - \sqrt{1-\sin^2 2\theta}\\…

Question 8(xix, xx, xxi & xxii) Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:46 +0000

Question 8(xix, xx, xxi & xxii) Exercise 8.2 Solutions of Question 8(xix, xx, xxi & xxii) of Exercise 8.2 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. $$\frac{\sin 2 \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\cos \alpha}=\sec \alpha$$\begin{align*} LHS &= \dfrac{\sin 2 \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\cos \alpha}\\ &= \dfrac{\sin 2 \alpha …

Question 1(i, ii, iii & iv) Exercise 8.3

anonymous@undisclosed.example.com (Anonymous) — Mon, 28 Oct 2024 17:14:30 +0000

Question 1(i, ii, iii & iv) Exercise 8.3 Solutions of Question 1(i, ii, iii & iv) of Exercise 8.3 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. $$4 \sin 16x \cos 10x $$\begin{align*} &4 \sin 16x \cos 10x \\ & = 2 (2\sin 16x \cos 10x) \\ &= 2[\sin(16x+10x)+\sin(16x-10x)]\\ &= 2[\sin (26x)+\sin(6x)] \end{align*}$10 \cos 10y \cos 6y$\begin{align*} &10 \…

Question 1(v, vi, vii & viii) Exercise 8.3

anonymous@undisclosed.example.com (Anonymous) — Mon, 28 Oct 2024 17:14:29 +0000

Question 1(v, vi, vii & viii) Exercise 8.3 Solutions of Question 1(v, vi, vii & viii) of Exercise 8.3 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(-u) \sin 5u$\begin{align*} &\sin(-u) \sin 5u \\ =& -\sin u \sin 5u \\ =& -\frac{1}{2}[\cos(u - 5u) - \cos(u + 5u)] \\ = &-\frac{1}{2}[\cos(-4u) - \cos(6u)] \\ =& \frac{1}{2}[\cos(6u) - \cos(4u) ] \e…

Question 9, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:41 +0000

Question 9, Exercise 8.1 Solutions of Question 9 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. $\alpha$$\beta$$\sin \alpha=\dfrac{1}{\sqrt{2}}$$\cos \beta=-\dfrac{3}{5}$$\sin (\alpha \pm \beta)$$\sin \alpha=\dfrac{1}{\sqrt{2}}$$\alpha$$\cos \beta=-\dfrac{3}{5}$$\beta$$$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$$\alpha$$\cos$\begin{align*…

Question 12, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:31 +0000

Question 12, Exercise 8.1 Solutions of Question 12 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. $\alpha+\beta+\gamma=180^{\circ}$$\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma$$$\alpha+\beta+\gamma=180^{\circ}$$\begin{align*} & \alpha+\beta=180^{\circ}-\gamma \\ \implies & \tan(\alpha+\beta) = \tan(180^{\circ}-…

Question 13, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:32 +0000

Question 13, Exercise 8.1 Solutions of Question 13 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. $r \sin (\theta+\phi)$$12 \sin \theta-5 \cos \theta$$12=r\cos \varphi $$-5=r\sin \varphi$\begin{align*} & (12)^2+(-5)^2=r^2 \cos^2\varphi+r^2 \sin^2 \varphi \\ \implies & 144+25={{r}^{2}}\left( {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi \r…

Question 1, 2 and 3 Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:35 +0000

Question 1, 2 and 3 Exercise 8.2 Solutions of Question 1, 2 and 3 of Exercise 8.2 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. $P(-3,4)$$\theta$$\theta$$\cos 2 \theta$$\sin 2 \theta$$2 \theta$$x=-3$$y=4$\begin{align*} r&= \sqrt{(-3)^2+4^2} \\ &=\sqrt{25} = 5. \end{align*}$$\sin\theta = \frac{4}{5} \text{ and } \cos\theta = -\frac{3}{5}.$$\begin{align…

Question 7 Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:47 +0000

Question 7 Exercise 8.2 Solutions of Question 7 of Exercise 8.2 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. $$\sin ^{2} \alpha \cos ^{2} \alpha$$\begin{align*} \sin ^{2} \alpha \cos ^{2} \alpha &= \left(\frac{1-\cos 2\alpha}{2} \right)\left(\frac{1+\cos 2\alpha}{2} \right)\\ &= \frac{1}{4}(1-\cos^2 2\alpha) \\ &=\frac{1}{4}\left(1-\frac{1+\cos 4\alp…

Question 8(i, ii & iii) Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:48 +0000

Question 8(i, ii & iii) Exercise 8.2 Solutions of Question 8(i, ii & iii) of Exercise 8.2 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. $(\sin \theta+\cos \theta)^{2}=1+\sin 2 \theta$\begin{align*} LHS & = (\sin \theta+\cos \theta)^{2} \\ &=\sin^2\theta + \cos^2\theta +2\sin \theta \cos\theta\\ &= 1+2\sin \theta \cos\theta \quad (\because \sin^2\theta…

Question 8(iv, v & vi) Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:50 +0000

Question 8(iv, v & vi) Exercise 8.2 Solutions of Question 8(iv, v & vi) of Exercise 8.2 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. $\csc 2 \alpha=\dfrac{\tan \alpha+\cot \alpha}{2}$\begin{align*} RHS & = \dfrac{\tan \alpha+\cot \alpha}{2} \\ & = \dfrac{1}{2}\left(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha} \right)\\ \end{align*}$8 \…

Question 8(vii, viii & ix) Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:53 +0000

Question 8(vii, viii & ix) Exercise 8.2 Solutions of Question 8(vii, viii & ix) of Exercise 8.2 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. $\sin 2 \theta=2 \cot \theta \sin ^{2} \theta$\begin{align*} RHS &= 2 \cot \theta \sin ^{2} \theta\\ &= 2 \frac{\cos \theta }{\sin \theta} \sin ^{2} \theta\\ &= 2 \cos \theta \sin\theta\\ &= \sin2 \theta\\ &=LH…

Question 8(x, xi & xii) Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:54:54 +0000

Question 8(x, xi & xii) Exercise 8.2 Solutions of Question 8(x, xi & xii) of Exercise 8.2 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. $\sec 2 x=\dfrac{\cos x}{\cos x+\sin x}+\dfrac{\sin x}{\cos x-\sin x}$\begin{align*} RHS &= \dfrac{\cos x}{\cos x+\sin x}+\dfrac{\sin x}{\cos x-\sin x}\\ &=\dfrac{\cos x(\cos x-\sin x)+\sin x(\cos x+\sin x)}{(\cos x+\…

Question 8(xiii, xiv & xv) Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:56:27 +0000

Question 8(xiii, xiv & xv) Exercise 8.2 Solutions of Question 8(xiii, xiv & xv) of Exercise 8.2 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. $\csc 2 \alpha-\cot 2 \alpha=\tan \alpha$\begin{align*} LHS &= \csc 2 \alpha-\cot 2 \alpha\\ &=\frac{1}{\sin 2 \alpha}- \frac{\cos2 \alpha}{\sin 2\alpha }\\ &=\frac{1-\cos2 \alpha}{\sin2 \alpha}\\ &= \frac{2\si…

Question 8(xvi, xvii & xviii) Exercise 8.2

anonymous@undisclosed.example.com (Anonymous) — Sat, 26 Oct 2024 18:56:02 +0000

Question 8(xvi, xvii & xviii) Exercise 8.2 Solutions of Question 8(xvi, xvii & xviii) of Exercise 8.2 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. $\dfrac{1-\cos ^{2} \beta}{2-2 \cos \beta}=\cos ^{2} \dfrac{\beta}{2}$\begin{align*} LHS &= \dfrac{1-\cos ^{2} \beta}{2-2 \cos \beta}\\ &= \dfrac{\sin ^{2} \beta}{2-2 \cos \beta}\\ &=\dfrac{4\sin ^{2} \fr…

Question 1(ix, x & xi) Exercise 8.3

anonymous@undisclosed.example.com (Anonymous) — Mon, 28 Oct 2024 17:14:30 +0000

Question 1(ix, x & xi) Exercise 8.3 Solutions of Question 1(ix, x & xi) of Exercise 8.3 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. $2 \sin 75{\circ} \sin 15{\circ}$\begin{align*} &\quad2 \sin 75^{\circ} \sin 15^{\circ} \\ &= \cos(75^{\circ} - 15^{\circ}) - \cos(75^{\circ} + 15^{\circ}) \\ &= \cos 60^{\circ} - \cos 90^{\circ} \\ \end{align*}$4 \sin …

Question 3(xi, xii & xiii) Exercise 8.3

anonymous@undisclosed.example.com (Anonymous) — Mon, 28 Oct 2024 17:14:32 +0000

Question 3(xi, xii & xiii) Exercise 8.3 Solutions of Question 3(xi, xii & xiii) of Exercise 8.3 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. $2\cos2u \cos u-\sin 2u \sin u=2\cos^3 u$\begin{align*} LHS & = 2\cos 2u \cos u - \sin 2u \sin u \\ & = 2\left(\cos^2 u - \sin^2 u\right) \cos u - 2\sin u \cos u \sin u \\ & = 2\cos^3 u - 2\sin^2 u \cos u \\ & =…

Question 4 Exercise 8.3

anonymous@undisclosed.example.com (Anonymous) — Mon, 28 Oct 2024 17:14:33 +0000

Question 4 Exercise 8.3 Solutions of Question 4 of Exercise 8.3 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. $\cos 80^{\circ} \cos 60^{\circ} \cos 40^{\circ} \cos 20^{\circ}=\dfrac{1}{16}$\begin{align*} LHS &= \cos 80^\circ \cos 60^\circ \cos 40^\circ \cos 20^\circ \\ &= \cos 80^\circ \left(\frac{1}{2}\right) \cos 40^\circ \cos 20^\circ \\ &= \frac{1…

Question 2, Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Sat, 09 Nov 2024 18:31:57 +0000

Question 2, Review Exercise Solutions of Question 2 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 \theta=\dfrac{3}{5}, \sin \phi=\dfrac{5}{13}$$\theta$$\phi$$\sin (\theta-\phi)$$\sin \theta=\dfrac{3}{5}$$\sin \phi=\dfrac{5}{13}$$\theta$$\phi$\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\frac{3}{5}\right)^2\\ & =…

Question 5 and 6, Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Sat, 09 Nov 2024 18:44:48 +0000

Question 5 and 6, Review Exercise Solutions of Question 5 and 6 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. $\tan \theta$$\tan \left(\theta-45^{\circ}\right)=\frac{1}{3}$\begin{align*} & \frac{\tan \theta - \tan 45^{\circ}}{1 + \tan \theta \cdot \tan 45^{\circ}} =\frac{1}{3}\\ \implies & \frac{\tan \theta - 1}{1 + \tan \theta}= \f…

Question 7, Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Sat, 09 Nov 2024 18:47:22 +0000

Question 7, Review Exercise Solutions of Question 7 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. $\dfrac{4 \sin ^{2} \theta \cos \theta}{\cos 3 \theta+\cos \theta}=\tan 2 \theta \tan \theta$\begin{align*} LHS&=\frac{4 \sin^2 \theta \cos \theta}{\cos 3 \theta + \cos \theta}\\ &=\frac{4 \sin \theta\sin \theta \cos \theta}{4\cos^ 3 \t…

Question 10, Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Sat, 09 Nov 2024 18:51:06 +0000

Question 10, Review Exercise Solutions of Question 10 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. $\sin (16 x)=16 \sin (x) \cos (x) \cos (2 x) \cos (4 x) \cos (8 x)$\begin{align*} RHS&=16 \sin (x) \cos (x) \cos (2 x) \cos (4 x) \cos (8 x) \\ &= 8(2 \sin (x) \cos (x) )\cos (2 x) \cos (4 x) \cos (8 x) \\ &= 8 \sin2 (x) \cos (2 x) \…

Question 5 and 6, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:39 +0000

Question 5 and 6, Exercise 8.1 Solutions of Question 5 and 6 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. $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$$\cos (\alpha+\beta)$$\cos (\alpha-\beta)$$\sin \alpha=\dfrac{4}{5}$$\alpha$$\tan \beta=-\dfrac{5}{12}$$\beta$$$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$$\alpha$$\cos$\begin{alig…

Question 14, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:54:47 +0000

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*} …

Question 3, Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Sat, 09 Nov 2024 18:32:33 +0000

Question 3, Review Exercise Solutions of Question 3 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. $\frac{1}{\sqrt{2}}(\sin \beta+\cos \beta)$\begin{align*} &\frac{1}{\sqrt{2}}(\sin \beta+\cos \beta)\\ =& \sin \frac{\pi}{4}\sin \beta+\cos \frac{\pi}{4}\cos \beta\\ =& \cos(\beta -\frac{\pi}{4}) \end{align*}$\frac{1}{\sqrt{2}} \sin 75^…

Question 4, Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Sat, 09 Nov 2024 18:33:14 +0000

Question 4, Review Exercise Solutions of Question 4 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. $\dfrac{1+\tan 15^{\circ}}{1-\tan 15^{\circ}}$\begin{align*} &\frac{1+\tan 15^{\circ}}{1-\tan 15^{\circ}}\\ =&\frac{1+\tan 15^{\circ}}{1-1 \cdot \tan 15^{\circ}}\\ =&\frac{\tan 45^{\circ} + \tan 15^{\circ}}{1 - \tan 45^{\circ} \tan 15^{…

Question 8, Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Sat, 09 Nov 2024 18:47:40 +0000

Question 8, Review Exercise Solutions of Question 8 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. $$\sqrt{\frac{\cos \left(90^{\circ}+x\right) \sec (-x) \tan \left(180^{\circ}-x\right)}{\sec \left(360^{\circ}-x\right) \sin \left(180^{\circ}+x\right) \cot \left(90^{\circ}-x\right)}}=i .$$\begin{align*} LHS&= \sqrt{\frac{\cos \left(90…

Question 7, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:39 +0000

Question 7, Exercise 8.1 Solutions of Question 7 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. $\alpha$$\beta$$\sin \alpha=\dfrac{12}{13}$$\tan \beta=\dfrac{4}{3}$$\sin(\alpha+\beta)$$\cos(\alpha+\beta)$$\tan(\alpha+\beta)$$\sin \alpha=\dfrac{12}{13}$$\alpha$$\tan \beta=\dfrac{4}{3}$$\beta$$$\cos \alpha=\pm \sqrt{1-\sin^2\alpha}.$$\(\a…

Question 8, Exercise 8.1

anonymous@undisclosed.example.com (Anonymous) — Sun, 20 Oct 2024 17:53:41 +0000

Question 8, Exercise 8.1 Solutions of Question 8 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. $\sin \alpha=\dfrac{3}{5}$$0<\alpha<\dfrac{\pi}{2}$$\cos \beta=\dfrac{12}{13}$$\dfrac{3 \pi}{2}<\beta<2 \pi$$\csc (\alpha+\beta)$$\sec (\alpha+\beta)$$\cot (\alpha+\beta)$$\sin \alpha=\dfrac{3}{5}$$0<\alpha<\dfrac{\pi}{2}$$\alpha$\begin{align…

Question 9, Review Exercise

anonymous@undisclosed.example.com (Anonymous) — Sat, 09 Nov 2024 18:49:13 +0000

Question 9, Review Exercise Solutions of Question 9 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. $$\sqrt{\frac{\left(1-\tan ^{2} x \cos (-x) \cos \left(360^{\circ}-x\right)\right) \tan 45^{\circ}}{\left\{\sin 90^{\circ}-\sin \left(180^{\circ}+x\right)\right\}\left\{\sin 90^{\circ}-\cos \left(90^{\circ}-x\right)\right\}}}$$\begin{al…