I am new to this website so Please forgive me for my mistakes.

I have a question of trigonometry to prove and it is as (I dont know how to write theta symbol sorry for it)

$(1-\sin \theta)/(1-\sec \theta) = 2\cot \theta(\cos \theta- \csc\theta)$

Thanks in advance!!!

  • $\begingroup$ Enclose the gray box inbetween \$\$ and type \theta (inside\$ \$) to get $\theta$. $\endgroup$ – Git Gud Nov 10 '13 at 15:33
  • 1
    $\begingroup$ ok thanx i am going to edit it $\endgroup$ – Bhuvnesh Gupta Nov 10 '13 at 15:34
  • $\begingroup$ Nice job. Now instead of $sin, sec, cot, cos$ and $cosec$, type \sin, \sec, \cot, \cos and \csc. $\endgroup$ – Git Gud Nov 10 '13 at 15:39
  • $\begingroup$ Are you sure you write the equations correctly. If you evalute Lhs and Rhs for instance for $\theta=pi/10$ you get different results. $\endgroup$ – Ömer Nov 10 '13 at 15:58
  • $\begingroup$ @Ömer I dont know whether question is correct or not but I have written it right.This question was asked in my exam $\endgroup$ – Bhuvnesh Gupta Nov 10 '13 at 16:00

Sometimes it helps to put everything in terms of $\sin\theta, \cos\theta$, first:

$$\frac{(1-\sin \theta)}{(1-\sec \theta)} = \frac{1-\sin \theta}{1 - \frac{1}{\cos\theta}} = \frac{(1 - \sin\theta)\cos\theta}{\cos\theta - 1} $$ $$\overset{?}= \quad2\cot \theta(\cos \theta- \csc\theta) = 2\frac{\cos \theta}{\sin\theta}\left(\cos \theta - \frac{1}{\cos\theta}\right) = (\sin(2\theta) - 2)\cot \theta \sec\theta$$

In truth, the first expression and the second expression, while perhaps describing similar functions over a small interval, are not equivalent expressions.

See Wolfram Alpha here, and see the graph below :

enter image description here

Now, if the question asked you to solve the equation: (to find values of $\theta$ for which the equation is true, this we can do). Indeed, the equality holds if and only if $$\theta = 2\pi n,\; n \in \mathbb Z\;\text{ or }\;\; \theta = 2\pi n - \frac \pi2, \;\;n \in \mathbb Z$$

  • $\begingroup$ OK I got it. thanx to explain in such a way $\endgroup$ – Bhuvnesh Gupta Nov 10 '13 at 16:39
  • $\begingroup$ You're very welcome, Bhuvnesh! $\endgroup$ – Namaste Nov 10 '13 at 16:40

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