# Question 5, Exercise 10.1

Solutions of Question 5 of Exercise 10.1 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

If $\tan \alpha =\dfrac{3}{4}$, $\sec \beta =\dfrac{13}{5}$ and neither the terminal side of the angle of measure $\alpha$ nor $\beta$ in the first Quadrant, then find: $\sin \left( \alpha +\beta \right)$.

Given: $\tan\alpha =\dfrac{3}{4}$.

As $\tan\alpha$ is +ive and terminal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \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 }\\ \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 $$\cos\alpha =\frac{1}{\sec \alpha }=\frac{1}{-\tfrac{5}{4}}$$ $$\Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align}\frac{\sin \alpha }{\cos \alpha }&=\tan \alpha \\ \Rightarrow \quad \sin \alpha &=\tan \alpha \cos \alpha \\ \Rightarrow \quad \sin \alpha &=\left(\frac{3}{4}\right)\left(-\frac{4}{5} \right)\\ \Rightarrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$

As \begin{align}\cos\beta &= \dfrac{1}{\sec\beta} = \dfrac{1}{\tfrac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align}

As $\cos\beta$ is +ive and terminal arm of $\beta$ is not in the 1st quadrant, it lies in 4th quadrant. Now \begin{align}\sin^{2}\beta &=1-\cos^2\beta \\ \Rightarrow \quad \sin\beta &=\pm\sqrt{1-\cos^2\beta}. \end{align} Since terminal ray of $\beta$ is in the fourth quadrant so value of $\sin$ is –ive, \begin{align}\sin\beta &=-\sqrt{1-{{\cos }^{2}}\beta } \\ \Rightarrow \quad \sin\beta &=-\sqrt{1-{{\left( \frac{5}{13} \right)}^{2}}}\\ &=-\sqrt{1-\frac{25}{169}}=-\sqrt{\frac{144}{169}} \\ \Rightarrow \quad \sin\beta&=-\dfrac{12}{13} \end{align}

Now \begin{align} \sin(\alpha+\beta)&=\sin\alpha \cos\beta +\cos\alpha \sin\beta \\ &=\left(-\frac{3}{5}\right)\left(\frac{5}{13}\right)+\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)\\ &=-\frac{3}{13}+\frac{48}{65}\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin(\alpha +\beta)=\dfrac{33}{65}.}$$

If $\tan \alpha =\dfrac{3}{4}$, $\sec \beta =\dfrac{13}{5}$ and neither the terminal side of the angle of measure $\alpha$ nor $\beta$ in the first Quadrant, then find: $\cos \left( \alpha +\beta \right)$.

Given: $\tan\alpha =\dfrac{3}{4}$.

As $\tan\alpha$ is +ive and terminal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \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 }\\ \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 $$\cos\alpha =\frac{1}{\sec \alpha }=\frac{1}{-\tfrac{5}{4}}$$ $$\Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align}\frac{\sin \alpha }{\cos \alpha }&=\tan \alpha \\ \Rightarrow \quad \sin \alpha &=\tan \alpha \cos \alpha \\ \Rightarrow \quad \sin \alpha &=\left(\frac{3}{4}\right)\left(-\frac{4}{5} \right)\\ \Rightarrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$

As \begin{align}\cos\beta &= \dfrac{1}{\sec\beta} = \dfrac{1}{\tfrac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align}

As $\cos\beta$ is +ive and terminal arm of $\beta$ is not in the 1st quadrant, it lies in 4th quadrant. Now \begin{align}\sin^{2}\beta &=1-\cos^2\beta \\ \Rightarrow \quad \sin\beta &=\pm\sqrt{1-\cos^2\beta}. \end{align} Since terminal ray of $\beta$ is in the fourth quadrant so value of $\sin$ is –ive, \begin{align}\sin\beta &=-\sqrt{1-{{\cos }^{2}}\beta } \\ \Rightarrow \quad \sin\beta &=-\sqrt{1-{{\left( \frac{5}{13} \right)}^{2}}}\\ &=-\sqrt{1-\frac{25}{169}}=-\sqrt{\frac{144}{169}} \\ \Rightarrow \quad \sin\beta&=-\dfrac{12}{13} \end{align}

Now \begin{align} \cos(\alpha+\beta)&=\cos\alpha \cos\beta -\sin\alpha \sin\beta \\ &=\left(-\frac{4}{5}\right)\left(\frac{5}{13}\right)-\left(\frac{3}{5}\right)\left(-\frac{12}{13}\right)\\ &=-\frac{20}{65}+\frac{36}{65}\end{align} $$\implies \bbox[4px,border:2px solid black]{\cos(\alpha +\beta)=\dfrac{16}{65}.}$$

If $\tan\alpha =\dfrac{3}{4}$, $\sec \beta =\dfrac{13}{5}$ and neither the terminal side of the angle of measure $\alpha$ nor $\beta$ in the first Quadrant, then find: $\tan \left( \alpha +\beta \right)$.

Given: $\tan\alpha =\dfrac{3}{4}$.

As $\tan\alpha$ is +ive and terminal arm of $\alpha$ in not in the 1st quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Rightarrow \quad \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 }\\ \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 $$\cos\alpha =\frac{1}{\sec \alpha }=\frac{1}{-\tfrac{5}{4}}$$ $$\Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align}\frac{\sin \alpha }{\cos \alpha }&=\tan \alpha \\ \Rightarrow \quad \sin \alpha &=\tan \alpha \cos \alpha \\ \Rightarrow \quad \sin \alpha &=\left(\frac{3}{4}\right)\left(-\frac{4}{5} \right)\\ \Rightarrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$

As \begin{align}\cos\beta &= \dfrac{1}{\sec\beta} = \dfrac{1}{\tfrac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align}

As $\cos\beta$ is +ive and terminal arm of $\beta$ is not in the 1st quadrant, it lies in 4th quadrant. Now \begin{align}\sin^{2}\beta &=1-\cos^2\beta \\ \Rightarrow \quad \sin\beta &=\pm\sqrt{1-\cos^2\beta}. \end{align} Since terminal ray of $\beta$ is in the fourth quadrant so value of $\sin$ is –ive, \begin{align}\sin\beta &=-\sqrt{1-{{\cos }^{2}}\beta } \\ \Rightarrow \quad \sin\beta &=-\sqrt{1-{{\left( \frac{5}{13} \right)}^{2}}}\\ &=-\sqrt{1-\frac{25}{169}}=-\sqrt{\frac{144}{169}} \\ \Rightarrow \quad \sin\beta&=-\dfrac{12}{13} \end{align} Now \begin{align} \tan(\alpha+\beta)&=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\ &= \dfrac{\sin\alpha \cos\beta -\cos\alpha \sin\beta}{\cos\alpha \cos\beta -\sin\alpha \sin\beta}\\ &= \dfrac{\left(-\frac{3}{5}\right)\left(\frac{5}{13}\right)+\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)}{\left(-\frac{4}{5}\right)\left(\frac{5}{13}\right)-\left(\frac{3}{5}\right)\left(-\frac{12}{13}\right)}\\ &=\dfrac{-\frac{3}{13}+\frac{48}{65}}{-\frac{20}{65}+\frac{36}{65}} &=\end{align} $$\implies \bbox[4px,border:2px solid black]{\tan(\alpha +\beta)=\dfrac{33}{16}.}$$

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