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To Find - $$\tan A + \cot A$$

Given, $$\sin A + \cos A = \sqrt2$$

My progress as far -

1st way- $$\Rightarrow \sin A = \sqrt2 - \cos A$$ $$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$ $$\Rightarrow \tan A = \frac{ \sqrt 2 }{\cos A } - 1$$ $$\Rightarrow \tan A + \cot A =\frac{ \sqrt 2 }{\cos A} -1 + \cot A $$

and the 2nd way as -

$$(\sin A + \cos A)^2 = 2$$ $$\sin ^2 A + \cos ^2 A + 2\sin A\cos A = 2 $$ $$\Rightarrow 2\sin A\cos A=1$$ $$\Rightarrow \sin A\cos A=\frac12$$

As we can see the first way is unable to give an answer in absolute Real Number, and the second way doesn't go even near to what is required to proof.

I know few trigonometry identities as per my textbook, those are

  • $\sin^2 A + \cos^2 A = 1$
  • $1 + \cot^2 A = \csc^2 A$
  • $\tan^2A + 1 = \sec^2 A$
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$$\tan{A}+\cot{A}=\frac{1}{\sin{A}\cos{A}}=\frac{2}{(\sin{A}+\cos{A})^2-1}=2$$

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  • $\begingroup$ didn't get your 3rd step $$\frac{2}{(sin A + cos A)^2 -1}$$ How was that? $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 4:23
  • $\begingroup$ @Abhas Kumar Sinha $\frac{1}{\sin{A}\cos{A}}=\frac{2}{2\sin{A}\cos{A}}=\frac{2}{\sin^2A+2\sin{A}\cos{A}+\cos^2A-1}=\frac{2}{(\sin{A}+\cos{A})^2-1}.$ $\endgroup$ – Michael Rozenberg Dec 31 '17 at 4:26
  • $\begingroup$ gotcha, quite difficult to do in a single step. $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 4:29
  • $\begingroup$ But it's exactly like I thought. $\endgroup$ – Michael Rozenberg Dec 31 '17 at 4:33
  • $\begingroup$ Can you solve this in my first way? $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 5:04
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Your second attempt is actually what helps here. Note:

$$\begin{align}\tan A+\cot A&=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\\&=\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}\\&=\frac{1}{\sin A\cos A}\\&=\frac{1}{1/2}\\&=2\end{align}$$

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  • $\begingroup$ how sinA.cosA = $ \frac{1}{2}$ ?? $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 4:27
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    $\begingroup$ Didn't you already derive that yourself, as your second attempt, which "doesn't go even near to what is required to proof". Umm... actually... it does! $\endgroup$ – user491874 Dec 31 '17 at 4:29
  • $\begingroup$ @AbhasKumarSinha when you did, $(\sin A+\cos A)^2=1+2\sin A\cos A$, you ended up with $\sin A\cos A=\dfrac12$, in your second approach :) $\endgroup$ – Awnon Bhowmik Dec 31 '17 at 4:29
  • $\begingroup$ @AbhasKumarSinha, hope you've managed to clarify it... $\endgroup$ – user491874 Dec 31 '17 at 4:41
  • $\begingroup$ @AwnonBhowmik yea, you've perfectly clarified everything, just questions how you are able to think this much in seconds? :) +1 upvote from me $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 4:43
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$$\begin{align}\tan A+\cot A&=\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}\\&=\dfrac{\sin^2A+\cos^2A}{\sin A\cos A}\\&=\dfrac1{\sin A\cos A}\\&=\dfrac2{2\sin A\cos A}\\&=\dfrac2{(\sin A+\cos A)^2-1}\\&=\dfrac2{(\sqrt2)^2-1}\\&=2\end{align}$$

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$\cos\frac {\pi}4=\sin\frac {\pi}4=\frac1{\sqrt{2}}$. Hence what you have got is $$\cos\frac {\pi}4\sin A+\sin\frac {\pi}4\cos A=1$$ or $\sin\Big(A+\frac {\pi}4\Big)=1$. Can you solve from here?

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  • $\begingroup$ $$\cos \frac {pi}{4} = \sin \frac{pi}{4}$$ Did't studied about it. $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 4:25
  • $\begingroup$ $\frac {\pi}4$ is same as $45^{\circ}$. Don't you know how to find the sine and cosine values for this angle? If not, note that the right angled triangle with one of its acute angle equal to $45^{\circ}$ is isosceles. $\endgroup$ – Abishanka Saha Dec 31 '17 at 4:37
  • $\begingroup$ how $ \frac{pi}{4} = 45$ degrees, pi is irrational and irrational / rational is always a irrational (also given if rational number is not takes = 0) $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 4:40
  • $\begingroup$ Their units are different. Sorry for not being more specific. Actually $\frac {\pi}4$ radians = $45$ degrees. $\endgroup$ – Abishanka Saha Dec 31 '17 at 4:42
  • $\begingroup$ Haven't studied radians, I'm 10th grade, $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 4:44
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Let's use $\sin^2A=\sqrt{1-\cos^2A}$

Equation transform into$$\sin A+\sqrt{1-\sin^2A}=\sqrt2$$

Which is quadratic equation and easy to solve

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  • $\begingroup$ what are you trying to say? I've to find tanA + cotA without using tables, only identities $\endgroup$ – Abhas Kumar Sinha Dec 31 '17 at 4:37
  • $\begingroup$ Find the value of sin A from the equation , then evaluate cos A using identity and then evaluate tanA +cotA $\endgroup$ – Atul Mishra Dec 31 '17 at 4:40

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