Question

If $\frac{\pi }{2}<\alpha <\pi ,\pi <\beta <\frac{3\pi }{2},\sin \left(\alpha \right)=\frac{15}{17},\tan \left(\beta \right)=\frac{12}{5}$, then the value of $\sin \left(\beta -\alpha \right)$ is

• A. $-\frac{171}{221}$
• B. $-\frac{21}{221}$
• C. $\frac{21}{221}$
• D. $\frac{171}{221}$

Answer 1

The correct option is D

$\frac{171}{221}$

Finding the value of $\sin \left(\beta -\alpha \right)$:

Step 1: Information required for the solution

The angle $\alpha$ is greater than $\frac{\pi }{2}$ and lesser than $\pi$ means it lies in second quadrant.

The angle $\beta$ is greater than $\pi$ and lesser than $\frac{3\pi }{2}$ means it lies in third quadrant.

To solve this we need the value of three trigonometric functions which are $\cos \left(\alpha \right),\sin \left(\beta \right),\cos \left(\beta \right),\sin \left(\alpha \right)$ so that we could use the trigonometric identity $\begin{array}{rcl}\mathbf{sin}\left(\beta -\alpha \right)& \mathbf{=}& \mathbf{sin}\left(\beta \right)\mathbf{cos}\left(\alpha \right)\mathbf{-}\mathbf{sin}\left(\alpha \right)\mathbf{cos}\left(\beta \right)\end{array}$.

Step 2: Calculation for the required trigonometric functions

Since $\sin \left(\alpha \right)=\frac{15}{17}$ then by the trigonometric identity ${\mathbf{sin}}^{\mathbf{2}}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}\mathbf{+}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}\mathbf{=}\mathbf{1}$, we have

$\begin{array}{rcl}\cos \left(\alpha \right)& =& \sqrt{1-{\sin }^2\left(\alpha \right)}\\ & =& \sqrt{1-{\left(\frac{15}{17}\right)}^2}\\ & =& \sqrt{\frac{289-225}{289}}\\ & =& \sqrt{\frac{64}{289}}\\ & =& -\frac{8}{17}\left[\because \alpha \text{is}\text{in}{2}^{\text{nd}}\text{quadrant}\right]\end{array}$

Now, we have $\tan \left(\beta \right)=\frac{12}{5}$ and $\mathbf{tan}\mathbf{\left(}\mathbf{\beta }\mathbf{\right)}\mathbf{=}\frac{\mathbf{\text{perpendicular}}}{\mathbf{\text{base}}}$ which means $\left(P\right)=12\text{units}$ and $\left(B\right)=5\text{units}$.

Here we will apply the Pythagoras theorem, ${\mathit{H}}^{\mathbf{2}}\mathbf{=}{\mathit{P}}^{\mathbf{2}}\mathbf{+}{\mathit{B}}^{\mathbf{2}}$, where $H$ is the hypotenuse.

$\begin{array}{rcl}H& =& \sqrt{{\left(12\right)}^2+{\left(5\right)}^2}\\ & =& \sqrt{144+25}\\ & =& \sqrt{169}\\ & =& 13\end{array}$

Then by the formulae $\mathbf{sin}\mathbf{\left(}\mathbf{\beta }\mathbf{\right)}\mathbf{=}\frac{\mathbf{P}}{\mathbf{H}}\text{and}\mathbf{cos}\mathbf{\left(}\mathbf{\beta }\mathbf{\right)}\mathbf{=}\frac{\mathbf{B}}{\mathbf{H}}$, we have

$\sin \left(\beta \right)=-\frac{12}{13}$

$\cos \left(\beta \right)=-\frac{5}{13}$

Values are negative because $\beta$ lies in the third quadrant.

Step 3: Determination of the value of $\mathbf{sin}\left(\beta -\alpha \right)$

By applying the identity, we have

$\begin{array}{rcl}\sin \left(\beta -\alpha \right)& =& \sin \left(\beta \right)\cos \left(\alpha \right)-\sin \left(\alpha \right)\cos \left(\beta \right)\\ & =& \left(-\frac{12}{13}\right)\left(-\frac{8}{17}\right)-\left(\frac{15}{17}\right)\left(-\frac{5}{13}\right)\\ & =& \frac{96}{221}+\frac{75}{221}\\ & =& \frac{96+75}{221}\\ & =& \frac{171}{221}\end{array}$

Hence, the correct option is (D).