1st way-$$Rightarrow sin A = sqrt2 - cos A$$$$Rightarrow an A = fracsqrt2 - cos Acos A$$$$Rightarrow an A = frac sqrt 2 cos A - 1$$$$Rightarrow an A + cot A =frac sqrt 2 cos A -1 + cot A $$

and the 2nd way as -

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

As we can see the first way is unable to lớn give an answer in absolute Real Number, & the second way doesn"t go even near khổng lồ what is required to proof.

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

$sin^2 A + cos^2 A = 1$$1 + cot^2 A = csc^2 A$$ an^2A + 1 = sec^2 A$
trigonometry proof-verification
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edited Dec 31, 2017 at 4:30
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Awnon Bhowmik
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asked Dec 31, 2017 at 4:17
user427802user427802
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$egingroup$
$$ anA+cotA=frac1sinAcosA=frac2(sinA+cosA)^2-1=2$$


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answered Dec 31, 2017 at 4:20
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Michael RozenbergMichael Rozenberg
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$egingroup$ didn't get your 3rd step $$frac2(sin A + cos A)^2 -1$$ How was that? $endgroup$
–user427802
Dec 31, 2017 at 4:23
$egingroup$ gotcha, quite difficult to vì in a single step. $endgroup$
–user427802
Dec 31, 2017 at 4:29
$egingroup$ Can you solve this in my first way? $endgroup$
–user427802
Dec 31, 2017 at 5:04
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$egingroup$
Your second attempt is actually what helps here. Note:

$$eginalign an A+cot A&=fracsin Acos A+fraccos Asin A\&=fracsin^2 A+cos^2 Asin Acos A\&=frac1sin Acos A\&=frac11/2\&=2endalign$$


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answered Dec 31, 2017 at 4:26
user491874user491874
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$egingroup$ how sinA.cosA = $ frac12$ ?? $endgroup$
–user427802
Dec 31, 2017 at 4:27


1
$egingroup$ Didn't you already derive that yourself, as your second attempt, which "doesn't go even near khổng lồ what is required lớn proof". Umm... Actually... It does! $endgroup$
–user491874
Dec 31, 2017 at 4:29
$egingroup$
AbhasKumarSinha, hope you've managed to clarify it... $endgroup$
–user491874
Dec 31, 2017 at 4:41
$egingroup$
AwnonBhowmik yea, you've perfectly clarified everything, just questions how you are able to lớn think this much in seconds? :) +1 upvote from me $endgroup$
–user427802
Dec 31, 2017 at 4:43
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$egingroup$
$$eginalign an A+cot A&=dfracsin Acos A+dfraccos Asin A\&=dfracsin^2A+cos^2Asin Acos A\&=dfrac1sin Acos A\&=dfrac22sin Acos A\&=dfrac2(sin A+cos A)^2-1\&=dfrac2(sqrt2)^2-1\&=2endalign$$


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answered Dec 31, 2017 at 4:22
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Awnon BhowmikAwnon Bhowmik
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$egingroup$
$cosfrac pi4=sinfrac pi4=frac1sqrt2$. Hence what you have got is $$cosfrac pi4sin A+sinfrac pi4cos A=1$$ or $sinBig(A+frac pi4Big)=1$. Can you solve from here?


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answered Dec 31, 2017 at 4:22
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QEDQED
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$egingroup$ $$cos frac pi4 = sin fracpi4$$ Did't studied about it. $endgroup$
–user427802
Dec 31, 2017 at 4:25


$egingroup$ how $ fracpi4 = 45$ degrees, pi is irrational và irrational / rational is always a irrational (also given if rational number is not takes = 0) $endgroup$
–user427802
Dec 31, 2017 at 4:40
$egingroup$ Haven't studied radians, I'm 10th grade, $endgroup$
–user427802
Dec 31, 2017 at 4:44
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$egingroup$
Let"s use $sin^2A=sqrt1-cos^2A$

Equation transform into$$sin A+sqrt1-sin^2A=sqrt2$$

Which is quadratic equation and easy to lớn solve


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answered Dec 31, 2017 at 4:32
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Atul MishraAtul Mishra
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$egingroup$ what are you trying lớn say? I've khổng lồ find tanA + cotA without using tables, only identities $endgroup$
–user427802
Dec 31, 2017 at 4:37
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