0
$\begingroup$

$$\sec2\theta=\csc2\theta$$

My attempt:

$$\begin{align} \cos2\theta &= \sin2\theta \tag{1}\\ \cos^2\theta+\sin^2\theta-2\cos\theta\sin\theta &=0 \tag{2}\\ (\cos\theta-\sin\theta)^2 &=0 \tag{3}\\ \cos\theta-\sin\theta &=0 \tag{4}\\ \cos\theta &=\sin\theta \tag{5}\\ \tan\theta &=1 \tag{6}\\ \theta &=180^\circ n+45^\circ \quad\text{??} \tag{7} \end{align}$$

But the answer was $90^\circ n+22.5^\circ$ and I'm not sure why. I've searched up the question online, and someone has proposed a solution where it is not factored; instead, the equation turns into $\tan2\theta=1$ on line $(2)$, and this allows you to get the correct solution.

What's wrong with factoring it though?

$\endgroup$
  • 3
    $\begingroup$ Second line, $\cos 2\theta = \cos^2\theta - \sin^2 \theta$. $\endgroup$ – Arthur Aug 3 '16 at 10:41
  • $\begingroup$ Noooooo. Ok thankyou $\endgroup$ – kjhg Aug 3 '16 at 10:41
  • 1
    $\begingroup$ BTW why don't you solve is as $\tan 2\theta=1$? $\endgroup$ – David Quinn Aug 3 '16 at 11:12
0
$\begingroup$

You have committed an error in the very beginning. Kindly note that

\begin{eqnarray} \cos(2\theta)=\cos^{2}(\theta)-\sin^{2}(\theta). \end{eqnarray}

You have written $\cos(2\theta)=\cos^{2}(\theta)+\sin^{2}(\theta)$, which is wrong.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.