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- Question 5, Exercise 10.1
- ad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alpha}.\end{align} Since terminal arm of $\alpha$ is in the ... ,}\\ \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now $$\cos\alpha =\frac{1}{\sec \alpha }=... t)\\ \Rightarrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$ As... frac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align} As $\cos\beta $ is +ive and terminal arm
- Question 13, Exercise 10.1
- ht) = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implie... cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{5}\end{align} and \begin{align}r\sin\varphi = 3 \implies... in\varphi = 3 \implies \sin\varphi =\dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\si... 3}{5}, \cos\varphi=\dfrac{4}{5} \text{ and } r=5.\end{align} =====Question 13(ii)===== Express each of
- Question 7, Exercise 10.2
- le identity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identi... quad (By \,using\, double\, angle\, identity)\end{align} =====Question 7(iii)===== Prove the ident... {2si{{n}^{2}}\theta }={{\cot }^{2}}\theta =R.H.S.\end{align} =====Question 7(iv)===== Prove the identi... }{2\sin \theta \cos \theta }=\tan \theta =R.H.S.\end{align} =====Question 7(v)===== Prove the identi
- Question 3, Exercise 10.1
- 25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$ As $... {9}{25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)... 2}{25}-\frac{12}{25}\\ \cos \left( u+v \right)&=0\end{align} =====Question 3(ii)===== If $\sin u=\df... \dfrac{4}{5},\,\,\,\,\,0\le v\le \dfrac{\pi }{2}.\end{align} We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u
- Question 1, Exercise 10.1
- =\sin \alpha \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\cir... \left( 37+22 \right) \\ & =\sin {{59}^{\circ }}. \end{align} ===== Question 1(ii)===== Write as a tr... &=\cos \alpha \cos \beta +\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{83}^{\circ ... cos \left( 83-53 \right)\\ &=\cos {{30}^{\circ }}\end{align}. ===== Question 1(iii)===== Write as a
- Question, Exercise 10.1
- }\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.... }}\\ \Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align} Now \begin{align}\sin (\alpha-\beta )&=\s... \cos \left( \alpha+\beta \right)&=\frac{33}{65}.\end{align} =====Question 4(ii)===== If $\sin \alpha... }\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.
- Question 2, Exercise 10.2
- (\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ ... dfrac{5}{13} \right)\left( \dfrac{12}{13} \right)\end{align} $$\implies \bbox[4px,border:2px solid blac... (\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ ... right)}^{2}}\\ &=\dfrac{144}{169}-\dfrac{25}{169}\end{align} $$\implies \bbox[4px,border:2px solid blac
- Question 2, Exercise 10.1
- \beta )=\sin \alpha \cos \beta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \lef... c{\sqrt{2}}{4} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4}. \end{align} ===Question 2(ii)=== Evaluate exactly:$... sqrt{3} \right)}{2}\\ &=2+\left( \sqrt{3} \right)\end{align} ===Question 2(iii)=== Evaluate exactly:$... 3}{-2}\\ &=\dfrac{4+\sqrt{3}}{-2}\\ &=-2-\sqrt{3}\end{align} ===Question 2(iv)=== Evaluate exactly:$\t
- Question11 and 12, Exercise 10.1
- a}{2}+\dfrac{\beta}{2}=90^\circ-\dfrac{\gamma}{2}\end{align} Now \begin{align}&\tan\left( \dfrac{\alpha... 90-\tfrac{\gamma}{2}\right)=\cot\tfrac{\gamma}{2}\end{align} \begin{align} \implies & \frac{\tan\dfrac{... pha}{2}\cot\dfrac{\beta}{2}\cot\dfrac{\gamma}{2}.\end{align} =====Question 12===== If $\alpha +\beta... irc\\ \implies &\alpha +\beta=180^\circ-\gamma. \end{align} Now \begin{align} & \tan \left( \alpha +\
- Question 6, Exercise 10.2
- ac{\sqrt{3}}{2}}{2}}=\dfrac{\sqrt{2+\sqrt{3}}}{2}\end{align} =====Question 6(ii)===== Use the half a... rt{\dfrac{2+1+2\sqrt{2}}{2-1}}=\sqrt{3+2\sqrt{2}}\end{align} =====Question 6(iii)===== Use the half... \sqrt{2}}{\sqrt{2}}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}\end{align} =====Question 6(iv)===== Use the half a... +1}{2\sqrt{2}}}\\ &=\dfrac{\sqrt{2+\sqrt{2}}}{2} \end{align} =====Question 6(v)===== Use the half an
- Question 8, Exercise 10.1
- eta +\sin\theta }{\cos\theta -\sin\theta }=R.H.S.\end{align} =====Question 8(ii)===== Prove that: $... right)}{\left( 1+\tan \theta \right)}\\ &=R.H.S.\end{align} ===Alternative Method=== \begin{align}L.... right)}{\left( 1+\tan \theta \right)}\\ &=R.H.S.\end{align} =====Question 8(iii)===== Prove that: ... tan }^{2}}\alpha {{\tan }^{2}}\beta }\\ &=R.H.S.\end{align} =====Question 8(iv)===== Prove that: $
- Question 4 and 5, Exercise 10.2
- frac{\pi}{2} < \frac{\theta}{2} < \dfrac{3\pi}{4}\end{align} This gives $\frac{\theta}{2}$ lies in 2nd ... right)}{2}}\\ &=-\sqrt{\dfrac{1+\dfrac{3}{7}}{2}}\end{align} $$\implies \bbox[4px,border:2px solid bla... c{\sqrt{3}}{2} \right)\left( \dfrac{1}{2} \right)\end{align} $$\implies \bbox[4px,border:2px solid blac... }{3}-1\\ &=2{{\left( \dfrac{1}{2} \right)}^{2}}-1\end{align} $$\implies \bbox[4px,border:2px solid blac
- Question 8 and 9, Exercise 10.2
- {8}\left[ 3+4\cos 2\theta +\cos 4\theta \right] \end{align} $$\implies \bbox[4px,border:2px solid blac... {3}}\theta -4\cos \theta \sin \theta \\ & = R.H.S\end{align} =====Question 9(ii)===== Prove the identi... c{1-{{\tan }^{2}}2\theta }{2\tan 2\theta }=R.H.S.\end{align} =====Question 9(iii)===== Prove the ident... ot^3\theta -3\cot\theta}{3\cot^2\theta -1}=R.H.S.\end{align} ====Go to ==== <text align="left"><btn typ
- Question 1, Exercise 10.3
- x\\ 2\sin 6x\sin x&=\cos 5\theta -\cos 7\theta\end{align} =====Question 1(ii)===== Express the pro... n {{178}^{\circ }}-\sin {{68}^{\circ }} \right]. \end{align} =====Question 1(iii)===== Express the pro... A+B}{2}-\dfrac{A-B}{2} \right)\\ &=\sin A+\sin B.\end{align} $$\implies \sin \dfrac{A+B}{2}\cos \dfrac{... {P+Q}{2}-\dfrac{P-Q}{2}\right)\\ &=\cos P-\cos Q \end{align} $$\implies\sin \dfrac{P+Q}{2}\cos \dfrac{P
- Question 2, Exercise 10.3
- ht)\cos \left( \dfrac{-{{6}^{\circ }}}{2} \right)\end{align} Since $cos(-\theta)=cos\theta$, we have $$... {2} \right).\\ &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, there... \ &=2\cos \dfrac{P}{2}\,\,\,\sin \dfrac{Q}{2}.\end{align} =====Question 2(iv)===== Convert the sum... \right)\\ &=2\cos \dfrac{A}{2} \cos\dfrac{B}{2}.\end{align} ====Go to==== <text align="left"><btn typ