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- Question 5, Exercise 10.1
- \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} \right)}^{2}}}=-\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{16}\,}\\ \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now $$\cos\alpha =\frac{1}{\sec \
- Question, Exercise 10.1
- &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1-{{\left(-\frac{4}{5} \right)}^{2}}}\\ &=-\sqrt{1-\dfrac{16}{25}}... dfrac{25}{169}}\\ \Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align} Now \begin{align}\sin (\alpha... pha\cos \beta+\cos \alpha\sin \beta \\ &=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{3}{5}\right)\left(\frac{5}{13}\right)\\ &=\dfrac{4
- Question 3, Exercise 10.1
- u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\sqrt{1-{{\left( \frac{3}{5} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{9}{25}}\,... sqrt{\dfrac{9}{25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos u\cos v-\sin u\sin v\\ &=\frac{4}{5}.\frac{3}{5}-\frac{3}{5}.\frac{4}{5}\\ &=\dfrac{12}{25}-\frac{12}{25}\\ \cos \left( u+v \right)&=0
- Question 2, Exercise 10.1
- gn} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\cos \frac{\pi }{3}\sin \frac{\pi }{4} \\ &=\left( \frac{\sqrt{3}}{2} \righ
- Question 2, Exercise 10.2
- \sqrt{1-{{\sin }^{2}}\theta }\\ &=-\sqrt{1-\left(\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, w... \sqrt{1-{{\sin }^{2}}\theta }\\ &=-\sqrt{1-\left(\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, w
- Question 4 and 5, Exercise 10.2
- ign}&\pi < \theta < \dfrac{3\pi}{2} \\ \implies &\frac{\pi}{2} < \frac{\theta}{2} < \dfrac{3\pi}{4}\end{align} This gives $\frac{\theta}{2}$ lies in 2nd quadrant. By using the h... ta}{2}=\pm\sqrt{\dfrac{1-\cos \theta }{2}}$$ As $\frac{\theta}{2}$ lies in 2nd quadrant and $\sin$ is po
- Question 8 and 9, Exercise 10.2
- \theta } \\ &= \dfrac{{{{\tan }^3}\theta \left( {\frac{1}{{{{\tan }^3}\theta }} - \frac{3}{{\tan \theta }}} \right)}}{{{{\tan }^3}\theta \left( {\frac{3}{{{{\tan }^2}\theta }} - 1} \right)}}\\ &=\dfr
- Question 1, Exercise 10.1
- beta },\end{align} Therefore \begin{align}\frac{\tan {{20}^{\circ }}+\tan {{32}^{\circ }}}{1-\tan... tion ==== As \begin{align}\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \bet
- Question 8, Exercise 10.1
- ( \dfrac{\pi }{4}-\theta \right)\\ &=\dfrac{\tan\frac{\pi }{4}-\tan\theta}{1+\tan\frac{\pi }{4}\tan\theta}\\ &=\dfrac{1-\tan\theta}{1+1\cdot\tan\theta } \qu
- Question 9 and 10, Exercise 10.1
- heta }{\cos ec4\theta }\\ &=\dfrac{\sin \theta }{\frac{1}{\cos 4\theta }}+\dfrac{\cos \theta }{\frac{1}{\sin 4\theta }}\\ &=\sin \theta \cos 4\theta +\cos \t
- Question 1, Exercise 10.2
- ta &=\dfrac{\sin 2\theta }{\cos 2\theta }=\dfrac{\frac{-5}{13}}{\frac{12}{13}}\end{align} $$ \implies \bbox[4px,border:2px solid black]{\tan 2\theta=-\dfrac{
- Question 3, Exercise 10.3
- =\dfrac{\cos 30^\circ}{\sin 30^\circ}\\ &=\dfrac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}=R.H.S.\end{align} =====Question 3(ii)===== Prove that $$\dfrac{\si
- Question11 and 12, Exercise 10.1
- c{\gamma}{2}\end{align} \begin{align} \implies & \frac{\tan\dfrac{\alpha}{2}\tan\dfrac{\beta}{2}\left(\d
- Question 1, Review Exercise 10
- ): $\dfrac{1}{2}$</collapse> vi. $\sin \left( x-\frac{\pi }{2} \right)=$ is * (a) $\sin x$