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- Question 1, Exercise 8.1
- on 1(i)===== Find the value of $\cos (\alpha \pm \beta), \sin (\alpha \pm \beta)$ and $\tan (\alpha \pm \beta)$ for the pair of angles. $\alpha=180^{\circ}, \beta=60^{\circ}$ ** Solution. ** Given: $\alpha=18
- Question 11, Exercise 8.1
- 270^{\circ}-\lambda\right)}=1$ ** Solution. ** \begin{align*} L.H.S & = \dfrac{\sin \left(180^{\circ... (90^{\circ}+\alpha\right)}=-1$ ** Solution. ** \begin{align*} L.H.S & = \frac{\sin \left(90^{\circ}+... estion 11(iii)===== Show that: $\tan \alpha+\tan \beta=\dfrac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}$ ** Solution. ** \begin{align*} L.H.S & = \ta
- Question 4 Exercise 8.2
- \theta$ lies in QI, $\sin$ is positive in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}... = \frac{4}{5} \end{align*} (a) $\sin 2 \theta$ \begin{align*} \sin 2\theta & = 2 \sin\theta \cos\the... \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}. \end{align*} (b) $\cos 2 \theta$ \begin{align*} \cos 2\theta & = 1-2\sin^2\theta \\ &
- Question 5 and 6, Exercise 8.1
- stion 5===== For $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$ with terminal side of an angles in QII, find $\cos (\alpha+\beta)$ and $\cos (\alpha-\beta)$. ** Solution. ** Given: $\sin \alpha=\dfrac{4}{5}$, $\alpha$ is in QII and $\tan \beta=-\dfrac{5}{12}$, $\beta$ is in QII. We have an
- Question 12, Exercise 8.1
- akistan. ===== Question 12(i)===== If $\alpha+\beta+\gamma=180^{\circ}$, prove that: $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma$. ** Solution. ** Given: $$\alpha+\beta+\gamma=180^{\circ}$$ This gives \begin{align*}
- Question 9, Exercise 8.1
- . ===== Question 9(i)===== Given $\alpha$ and $\beta$ are obtuse angles with $\sin \alpha=\dfrac{1}{\sqrt{2}}$ and $\cos \beta=-\dfrac{3}{5}$ find: $\sin (\alpha \pm \beta)$ ** Solution. ** Given: $\sin \alpha=\dfrac{1}{\sq... pha$ is obtuse angle, i.e. it is in QII.\\ $\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. i
- Question 7, Exercise 8.1
- an. ===== Question 7===== Given $\alpha$ and $\beta$ are acute angles with $\sin \alpha=\dfrac{12}{13}$ and $\tan \beta=\dfrac{4}{3}$ find \\ (i) $\sin(\alpha+\beta)$ (ii) $\cos(\alpha+\beta)$ (iii) $\tan(\alpha+\beta)$. ** Solution. ** Given: $\sin \alpha=\dfra
- Question 8, Exercise 8.1
- }{5}$, where $0<\alpha<\dfrac{\pi}{2}$ and $\cos \beta=\dfrac{12}{13}$, where $\dfrac{3 \pi}{2}<\beta<2 \pi$ find: \\ (i) $\csc (\alpha+\beta)$ (ii) $\sec (\alpha+\beta)$ (iii) $\cot (\alpha+\beta)$ ** Solution. ** Given: $\sin \alpha=\df
- Question 3(vi, vii, viii, ix & x) Exercise 8.3
- s 3y= \sec y(\sin 4y-\sin 2y)$ ** Solution. ** \begin{align*} LHS & = 2\tan y \cos 3y \\ & = 2 \cdot... o 3(vii)===== Prove the identity $\dfrac{ \sin 6 \beta + \sin 4 \beta}{\sin 6 \beta - \sin 4 \beta}=\tan 5 \beta \cot \beta$ ** Solution. ** \begin{align*} LHS & = \df
- Question 5 Exercise 8.2
- $ in QII, therefore $\cos 2\theta$ is negative. \begin{align*}\cos 2\theta & = - \sqrt{1-\sin^2 2\the... $\theta$ lies in QI and $\sin\theta > 0$. Thus \begin{align*} \sin\theta & = \sqrt{\frac{1-\cos 2\th... 32}{25}}{2}} = \sqrt{\frac{16}{25}} \end{align*} \begin{align*} \implies \boxed{\sin\theta = \frac{4}{... eta$ lies in QI, therefore $\cos\theta >0$, thus \begin{align*} \cos\theta & = \sqrt{1-\sin\theta} \\
- Question 3(i, ii, iii, iv & v) Exercise 8.3
- )===== Prove the identity $\dfrac{\cos (\alpha + \beta)}{\cos(\alpha - \beta)}=\dfrac{1- \tan \alpha \tan \beta}{1+ \tan \alpha \tan \beta}$ ** Solution. ** \begin{align*} RHS & = \dfrac{1- \tan \alpha \tan \b
- Question 8(x, xi & xii) Exercise 8.2
- \dfrac{\sin x}{\cos x-\sin x}$ ** Solution. ** \begin{align*} RHS &= \dfrac{\cos x}{\cos x+\sin x}+\... s ^{4} x-\sin ^{4} x=\cos 2 x$ ** Solution. ** \begin{align*} LHS &= \cos ^{4} x-\sin ^{4} x\\ &=(\c... n 8(xii)===== Verify the identities: $\tan \frac{\beta}{2}+\cot \frac{\beta}{2}=2 \csc \beta$ ** Solution. ** \begin{align*} RHS &= \tan \frac{\beta}{2}
- Question 8(xvi, xvii & xviii) Exercise 8.2
- ===== Verify the identities: $\dfrac{1-\cos ^{2} \beta}{2-2 \cos \beta}=\cos ^{2} \dfrac{\beta}{2}$ ** Solution. ** \begin{align*} LHS &= \dfrac{1-\cos ^{2} \beta}{2-2 \cos \beta}\\ &= \dfrac{
- Question 8(xix, xx, xxi & xxii) Exercise 8.2
- ha}{\cos \alpha}=\sec \alpha$$ ** Solution. ** \begin{align*} LHS &= \dfrac{\sin 2 \alpha}{\sin \alp... xx)===== Verify the identity: $2 \sin ^{2} \frac{\beta}{2}+\cos \beta=1$ ** Solution. ** \begin{align*} LHS & = 2 \sin ^{2} \frac{\beta}{2}+\cos \beta \\ & = 2 \sin^2
- Question 2, Review Exercise
- where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1... theta <0$, thus $$\cos \theta=-\frac{4}{5}$$ Now \begin{align*} \cos^2 \phi &= 1-\sin^2\phi\\ &= 1-\le... 0$, thus $$\cos \phi=\frac{12}{13}$$ As, we have \begin{align*} \sin(\theta -\phi)&=\sin \theta \cos \... where $\theta$ is obtuse and $\phi$ is acute. As\begin{align*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1