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- Question 5, Exercise 10.1
- nt, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Righ... sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alpha}.\end{align} Since terminal arm of $\alpha$ is in the 3rd quadrant, value of $\sec$ is –ive \begin{align}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Ri... \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now $$\cos\alpha =\frac{1}{\sec \alpha }=\frac{
- Question 13, Exercise 10.1
- nd $3=r\sin\varphi$. Squaring and adding \begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+3^2\\ ... = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{5}\end{align} and \begin{align}r\sin\varphi = 3 \implies 5\sin
- Question 7, Exercise 10.2
- dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\sin }^{4}}\theta \... dentity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identity $\t... \dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+co\operatorname{t}... (By \,using\, double\, angle\, identity)\end{align} =====Question 7(iii)===== Prove the identity $\
- Question 3, Exercise 10.1
- t and $\cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2... \\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$ As $v$ lie... nt and $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2... 25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos
- Question 1, Exercise 10.1
- n {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22}^{\circ }}+\cos {{37}^{\circ }}\sin {{22}^{\cir... t( 37+22 \right) \\ & =\sin {{59}^{\circ }}. \end{align} ===== Question 1(ii)===== Write as a trigonom
- Question, Exercise 10.1
- adrant and $\cos$ is -ive in 3rd quadrant, \begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1... \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.$$ As ... adrant and $\sin$ is +ive in 2nd quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\... \Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align} Now \begin{align}\sin (\alpha-\beta )&=\sin \al
- Question 2, Exercise 10.2
- cond quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }\\ &=-... ac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, ... ollowing by using double angle identities: \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ &=2\le... c{5}{13} \right)\left( \dfrac{12}{13} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\si
- Question 2, Exercise 10.1
- }-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{... qrt{2}}{4} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4}. \end{align} ===Question 2(ii)=== Evaluate exactly:$\tan {
- Question11 and 12, Exercise 10.1
- $\gamma$ are angles of triangle, therefore \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implies &\al... }+\dfrac{\beta}{2}=90^\circ-\dfrac{\gamma}{2}\end{align} Now \begin{align}&\tan\left( \dfrac{\alpha }{2}+\dfrac{\beta }{2}\right)=\tan \left( 90-\dfrac{\gamma }... tfrac{\gamma}{2}\right)=\cot\tfrac{\gamma}{2}\end{align} \begin{align} \implies & \frac{\tan\dfrac{\alpha
- Question 6, Exercise 10.2
- {\circ }}$by using half angle identity as, \begin{align}\cos {{15}^{\circ }}&=\cos \dfrac{{{30}^{\circ }}... sqrt{3}}{2}}{2}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}\end{align} =====Question 6(ii)===== Use the half angle i... {\circ }}$by using half angle identity as, \begin{align}\tan {{67.5}^{\circ }}&=\tan \dfrac{{{135}^{\circ... dfrac{2+1+2\sqrt{2}}{2-1}}=\sqrt{3+2\sqrt{2}}\end{align} =====Question 6(iii)===== Use the half angle
- Question 8, Exercise 10.1
- \theta -\sin \theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\theta \righ... +\sin\theta }{\cos\theta -\sin\theta }=R.H.S.\end{align} =====Question 8(ii)===== Prove that: $\tan \... \theta }{1+tan\theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}-\theta \righ... t)}{\left( 1+\tan \theta \right)}\\ &=R.H.S.\end{align} ===Alternative Method=== \begin{align}L.H.S.&=
- Question 4 and 5, Exercise 10.2
- theta$ is in the third quadrant, that is, \begin{align}&\pi < \theta < \dfrac{3\pi}{2} \\ \implies &\fra... {\pi}{2} < \frac{\theta}{2} < \dfrac{3\pi}{4}\end{align} This gives $\frac{\theta}{2}$ lies in 2nd quadra... in$ is positive in 2nd quadrant, therefore \begin{align}\sin\dfrac{\theta }{2}&=-\sqrt{\dfrac{1-\cos \the... t)}{2}}\\ &=-\sqrt{\dfrac{1+\dfrac{3}{7}}{2}}\end{align} $$\implies \bbox[4px,border:2px solid black]{\s
- Question 2, Exercise 10.3
- {{37}^{\circ }}$, $\beta ={{43}^{\circ }}$ \begin{align}\sin {{37}^{\circ }}+\sin {{43}^{\circ }}&=2\sin ... cos \left( \dfrac{-{{6}^{\circ }}}{2} \right)\end{align} Since $cos(-\theta)=cos\theta$, we have $$\sin {... $\alpha =36^\circ$, adn $\beta =82^\circ$ \begin{align}\cos {{36}^{\circ }}-\cos {{82}^{\circ }}&=-2\sin... \right).\\ &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $
- Question 8 and 9, Exercise 10.2
- or more cosine functions. ====Solution==== \begin{align}{{\cos}^{4}}\theta &={{\left( {{\cos }^{2}}\theta... left[ 3+4\cos 2\theta +\cos 4\theta \right] \end{align} $$\implies \bbox[4px,border:2px solid black]{\co... in \theta \cos \theta $ . ====Solution==== \begin{align}L.H.S.&=\sin 4\theta \\ &=\sin 2\left( 2\theta \... \theta -4\cos \theta \sin \theta \\ & = R.H.S\end{align} =====Question 9(ii)===== Prove the identity $\c
- Question 1, Exercise 10.3
- ta ).$$ Put $\alpha =6x$ and $\beta =x$ \begin{align}-\,2\sin 6x\sin x&=\cos (6x+x)-\cos (6x-x)\\ &=\c... 2\sin 6x\sin x&=\cos 5\theta -\cos 7\theta\end{align} =====Question 1(ii)===== Express the product a... {\circ }}$ and $\beta ={{123}^{\circ }}$ \begin{align}2\sin {{55}^{\circ }}\cos {{123}^{\circ }}&=\sin ... 178}^{\circ }}-\sin {{68}^{\circ }} \right]. \end{align} =====Question 1(iii)===== Express the product a