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- Question 5, Exercise 10.1
- lign}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alph... gn}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} ... -\sqrt{1+\frac{9}{16}}=-\sqrt{\frac{25}{16}\,}\\ \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now ... }{\sec \alpha }=\frac{1}{-\tfrac{5}{4}}$$ $$\Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align
- Question 3, Exercise 10.1
- \cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2}}u}\\ &=\s... qrt{1-\dfrac{9}{25}}\,\,=\sqrt{\dfrac{16}{25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\... $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2}}v}\\ &=\s... qrt{1-\dfrac{16}{25}}\,\,=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begi
- Question, Exercise 10.1
- =-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also... qrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}\\ \Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align} Now ... 5}{13}\right)\\ &=\dfrac{48}{65}-\frac{15}{65}\\ \Rightarrow \quad \cos \left( \alpha+\beta \right)&=\frac{33}... =-\sqrt{1-\dfrac{16}{25}}=\sqrt{\dfrac{9}{25}}\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also
- Question 2, Exercise 10.1
- eta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi }{4} \... 0)&=\dfrac{\tan 45+\tan 30}{1-\tan 45\tan 30}.\\ \Rightarrow \,\,\tan (75)&=\dfrac{1+\dfrac{1}{\sqrt{3}}}{1-\l