If $$\frac{\cos\theta}{\cos\alpha}+\frac{\sin\theta}{\sin\alpha}=\frac{\cos\delta}{\cos\alpha}+\frac{\sin\delta}{\sin\alpha}=1,$$ where $\theta$ and $\delta$ do not differ by an even multiple of $\pi$, then prove that $$\frac{\cos\theta\cos\delta}{\cos^2\alpha}+\frac{\sin\theta\sin\delta}{\sin^2\alpha}+1=0.$$

$$\frac{\cos\theta}{\cos\alpha}+\frac{\sin\theta}{\sin\alpha}=\frac{\cos\delta}{\cos\alpha}+\frac{\sin\delta}{\sin\alpha}=1$$ $$\frac{\cos\theta \sin\alpha+\sin\theta\cos\alpha}{\sin\alpha\cos\alpha}=\frac{\cos\delta \sin\alpha+\sin\delta\cos\alpha}{\sin\alpha\cos\alpha}=1$$
We need to prove that $$\frac{\cos\theta\cos\delta}{\cos^2\alpha}+\frac{\sin\theta\sin\delta}{\sin^2\alpha}+1=0$$ $$\implies\frac{\cos\theta\cos\delta-\cos^2\alpha\cos(\theta+\delta)}{\sin^2\alpha\cos^2\alpha}+1=0$$
I am stuck here. Please help.

  • 2
    $\begingroup$ So, we have $\sin(\theta+\alpha)=\sin(\alpha+\delta)=\sin\alpha\cos\alpha$ $\implies \theta+\alpha=(2n+1)\pi-(\alpha+\delta)$ $\endgroup$ – lab bhattacharjee May 5 '16 at 14:19
  • 2
    $\begingroup$ Have you tried multiplying $\frac{\cos\theta}{\cos\alpha}+\frac{\sin\theta}{\sin\alpha}$ and $\frac{\cos\delta}{\cos\alpha}+\frac{\sin\delta}{\sin\alpha}$ ? This will be equal to 1. You will have the two terms you wanted plus two other terms that sums up to 2 I think. $\endgroup$ – H. Potter May 5 '16 at 14:21
  • $\begingroup$ I got the answer by the above two useful comments.Thanks a lot. $\endgroup$ – Vinod Kumar Punia May 5 '16 at 14:27

For future readers, I here combine the comments by @H.Potter and @lab bhattacharjee into an answer.

Let $\frac{\cos \theta}{\cos\alpha} $ be denoted as $(1)$,

$\frac{\sin \theta}{\sin \alpha} $ be denoted as $(2)$,

$\frac{\cos \delta}{\cos\alpha} $ be denoted as $(3)$, and

$\frac{\sin \delta}{\sin\alpha} $ be denoted as $(4)$.

As given in the question, $$(1) +(2) = (3) +(4) = 1$$

Mupltiplying $\big((1) + (2) \big)$ by $\big((3) + (4)\big)$ :

$$\left(\frac{\cos \theta}{\cos\alpha} + \frac{\sin \theta}{\sin \alpha}\right)\left(\frac{\cos \delta}{\cos\alpha} + \frac{\sin \delta}{\sin\alpha}\right) =1 \times 1$$

Expanding the left hand side :

$$\frac{\cos \theta \, \cos \delta}{\cos^2\alpha} + \frac{\sin \theta \, \sin \delta}{\sin^2\alpha} + \frac {2\sin \theta \cos \delta }{\sin\alpha \cos \alpha} = 1 \,\,\,(♣)$$

As you can hopefully see, the first two fractions are the ones as required in the final answer/"hence proved step". So, we just need to make the third fraction equal to $2$ so that when the $1$ of $RHS$ comes to $LHS$, it completes the proof.

Now , let us focus on the original statement :

$(1) + (2) = (3)+(4) = 1$

As OP has already mentioned in the "Question Details" section :

$$\frac {\cos \theta \sin \alpha + \sin \theta \cos \alpha} {\sin \alpha \cos \alpha} = \frac {\cos \delta \sin \alpha + \sin \delta \cos \alpha}{\sin \alpha \cos \alpha} = 1 $$

$$\implies \frac {\sin (\theta + \alpha)}{\sin \alpha \cos \alpha} = \frac {\sin (\alpha + \delta)}{\sin \alpha \cos \alpha} = 1$$

Now you can equate $\sin (\theta + \alpha)$ and $\sin (\alpha + \delta)$, giving $\theta + \alpha = (2n+1)\pi - (\alpha + \delta)$.

Hopefully, this is enough to conclude that $\large \frac {2\sin \theta \cos \delta}{\sin \alpha \cos \alpha} = 2. $

Plug this back in (♣) to complete the proof.

Hope this helps :).


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