Question

If (a-b) sin ($\theta +\varphi$) = (a+b) sin ($\theta +\varphi$) and a tan$\left(\frac{\theta }{2}\right)$ - btan$\left(\frac{\varphi }{2}\right)$ = C,the the value of sin$\varphi$ is equal

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We want to find sin $\varphi$. We can find sin $\varphi$, if we know cos $\varphi$ or tan $\varphi$ or tan $\frac{\varphi }{2}$ or cos2$\varphi$. [There are other ratios which will help us in finding sin $\varphi$].

We will try to simplify / modify the given expressions with the hope that $\theta$ it will help us in finding any of the above ratios (a-b) sin ($\theta$-$\varphi$) = (a+b) sin ($\theta$-$\varphi$)

This is one of the expressions given to us. We solved similar problems and saw that one of the ways to proceed is by applying componendo - dividendo.

$\Rightarrow$ $\frac{a-b}{a+b}$ = $\frac{sin(\theta -\varphi )}{sin(\theta +\varphi )}$

$\Rightarrow$ $\frac{a-b+a+b}{a+b-(a-b)}$ = $\frac{sin(\theta -\varphi )+sin(\theta +\varphi )}{sin(\theta +\varphi )-sin(\theta -\varphi )}$

$\frac{a}{b}$ = $\frac{2sin\theta cos\varphi }{2cos\theta sin\varphi }$

$\Rightarrow$ $\frac{a}{b}$ = $\frac{tan\theta }{tan\varphi }$

$\Rightarrow$ tan$\theta$ = $\frac{a}{b}$ tan$\varphi$ ___________(1)

Since we want to find sin$\varphi$,this expression helps us to eliminate $\theta$ with expression

a tan$\frac{\theta }{2}$ - b tan$\frac{\varphi }{2}$ = C

$\Rightarrow$ $tan\frac{\theta }{2}$ = $\frac{c+btan\frac{\varphi }{2}}{a}$ ______(2)

We know tan$\theta$ = $\frac{2tan\frac{\theta }{2}}{1-ta{n}^2\frac{\theta }{2}}$

We will use this to eliminate tan$\theta$ and $\tan \frac{\theta }{2}$ with (1) and (2)

$\Rightarrow$ 2 $\times \frac{\frac{c+btan\frac{\varphi }{2}}{a}}{1-{\left(\frac{c+btan\frac{\varphi }{2}}{a}\right)}^2}=\frac{a}{b}tan\varphi$

After seeing the above relation, we can guess that we will be able to find tan$\frac{\varphi }{2}$ from it. Using tan$\frac{\varphi }{2}$ we can find sin$\varphi$,

because sin$\varphi$ = $\frac{2tan\frac{\varphi }{2}}{1+ta{n}^2\frac{\varphi }{2}}$
= $\frac{2a(c+btan\frac{\varphi }{2})}{a^2-{(c+btan\frac{\varphi }{2})}^2}=\frac{a}{b}\times \frac{2tan\frac{\varphi }{2}}{1-ta{n}^2\frac{\varphi }{2}}$

$\Rightarrow$ bc + $b^{2}tan\frac{\varphi }{2}$ - bc $ta{n}^2\frac{\varphi }{2}$ - $b^{2}ta{n}^3\frac{\varphi }{2}$ = $a^{2}tan\frac{\varphi }{2}$ - $c^{2}tan\frac{\varphi }{2}$ - $b^{2}ta{n}^3\frac{\varphi }{2}$ - $2bcta{n}^2\frac{\varphi }{2}$

$\Rightarrow$ bc + $(b^2-a^2+c^2)tan\frac{\varphi }{2}$ = -bc $ta{n}^2\frac{\varphi }{2}$

$\Rightarrow$ bc + $(1+ta{n}^2\frac{\varphi }{2})$ = $tan\frac{\varphi }{2}(a^2-b^2-c^2)$

$\Rightarrow$ $\frac{tan\frac{\varphi }{2}}{1+ta{n}^2\frac{\varphi }{2}}$ = $\frac{bc}{(a^2-b^2-c^2)}$

$\Rightarrow$ $sin\varphi$ = $\frac{2tan\frac{\varphi }{2}}{1+ta{n}^2\frac{\varphi }{2}}$ = $\frac{2bc}{(a^2-b^2-c^2)}$

[In the last four steps, we tried to get $\frac{tan\frac{\varphi }{2}}{1+ta{n}^2\frac{\varphi }{2}}$ because we see the possibilities of finding it]