Question

If $\sin \theta +\sin \varphi =a$ and $\cos \theta +\cos \varphi =b$, then $\tan \left(\frac{\theta -\varphi }{2}\right)$ is equal to

• A. $\sqrt{\frac{\left(a^2+b^2\right)}{\left(4-a^2-b^2\right)}}$
• B. $\sqrt{\frac{\left(4-a^2-b^2\right)}{\left(a^2+b^2\right)}}$
• C. $\sqrt{\frac{\left(a^2+b^2\right)}{\left(4+a^2+b^2\right)}}$
• D. $\sqrt{\frac{\left(4+a^2+b^2\right)}{\left(a^2+b^2\right)}}$

Answer 1

The correct option is B

$\sqrt{\frac{\left(4-a^2-b^2\right)}{\left(a^2+b^2\right)}}$

Explanation for the correct answer

Step 1: Conversion of the angle as $\left(\theta -\varphi \right)$

We know from the given that

$a=\sin \theta +\sin \varphi$

Square both the sides,

$\begin{array}{rcl}\therefore a^2& =& {\left(\sin \theta +\sin \varphi \right)}^2\\ & \Rightarrow & a^2={\sin }^2\theta +{\sin }^2\varphi +2\sin \theta \sin \varphi \dots \left(1\right)\end{array}$

and

$b=\cos \theta +\cos \varphi$

For this also, square both the sides,

$\begin{array}{rcl}\therefore b^2& =& {\left(\cos \theta +\cos \varphi \right)}^2\\ & \Rightarrow & b^2={\cos }^2\theta +{\cos }^2\varphi +2\cos \theta \cos \varphi \dots \left(2\right)\end{array}$

Add, both the equations $\left(1\right)\text{and}\left(2\right)$,

$\begin{array}{rcl}{a}^2+b^2& =& \sin {}^2\theta +\sin {}^2\varphi +2\sin \theta \sin \varphi +{\cos }^2\theta +{\cos }^2\varphi +2\cos \theta \cos \varphi \\ & =& \left({\sin }^2\theta +{\cos }^2\theta \right)+\left({\sin }^2\varphi +{\cos }^2\varphi \right)+2\left(\sin \theta \sin \varphi +\cos \theta \cos \varphi \right)\end{array}$

Here, two trigonometric identities will be applied which are ${\mathbf{sin}}^{\mathbf{2}}\left(A\right)\mathbf{+}{\mathbf{cos}}^{\mathbf{2}}\left(A\right)\mathbf{=}\mathbf{1}\mathbf{}\text{and}\mathbf{cos}\left(\theta -\varphi \right)\mathbf{=}\mathbf{cos}\left(\theta \right)\mathbf{cos}\left(\varphi \right)\mathbf{+}\mathbf{sin}\left(\theta \right)\mathbf{sin}\left(\varphi \right)$ then the equation will become,

$\begin{array}{rcl}{a}^2+b^2& =& 1+1+2\left(\cos \left(\theta -\varphi \right)\right)\\ & =& 2+2\left(\cos \left(\theta -\varphi \right)\right)\\ & =& 2\left(1+\cos \left(\theta -\varphi \right)\right)\end{array}$

Step 2: Determination of the required

From the above step, we know that

$\begin{array}{rcl}1+\cos \left(\theta -\varphi \right)& =& \frac{a^2+b^2}{2}\dots \left(3\right)\end{array}$

We know that $\mathbf{cos}\mathbf{\left(}\mathbf{2}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{2}{\mathbf{cos}}^{\mathbf{2}}\mathit{A}\mathbf{-}\mathbf{1}$ then $2{\cos }^{2}A=1+\cos \left(2\mathrm{A}\right)$

Similar to the half-angle formula: $2{\cos }^2\left(\frac{A}{2}\right)=1+\cos \left(\mathrm{A}\right)$

Compare this with the equation $\left(3\right)$,

$\begin{array}{rcl}\therefore 2{\cos }^2\left(\frac{\theta -\varphi }{2}\right)& =& \frac{a^2+b^2}{2}\\ & \Rightarrow & {\cos }^2\left(\frac{\theta -\varphi }{2}\right)=\frac{a^2+b^2}{4}\dots \left(4\right)\end{array}$

Step 3: Apply the identity ${\mathbf{tan}}^{\mathbf{2}}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}{\mathbf{sec}}^{\mathbf{2}}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{-}\mathbf{1}$, then

$\begin{array}{rcl}\therefore {\tan }^2\left(\frac{\theta -\varphi }{2}\right)& =& {\sec }^2\left(\frac{\theta -\varphi }{2}\right)-1\\ & =& \frac{1}{{\cos }^2\left(\frac{\theta -\varphi }{2}\right)}-1\left[\because \sec \left(A\right)=\frac{1}{\cos \left(A\right)}\right]\\ & =& \frac{1}{{\displaystyle \frac{a^2+b^2}{4}}}-1\left[\text{from}\left(4\right)\right]\\ & =& \frac{4}{a^2+b^2}-1\\ & =& \frac{4-a^2-b^2}{a^2+b^2}\end{array}$

From this, we have

$\tan \begin{array}{r}\left(\frac{\theta -\varphi }{2}\right)\end{array}=\sqrt{\frac{4-a^2-b^2}{a^2+b^2}}$

Hence, the correct option is (B).