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

Awnon Bhowmik

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asked Dec 31, 2017 at 4:17

user427802user427802

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## 5 Answers 5

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2

$egingroup$

$$ anA+cotA=frac1sinAcosA=frac2(sinA+cosA)^2-1=2$$

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answered Dec 31, 2017 at 4:20

Michael RozenbergMichael Rozenberg

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19

$egingroup$ 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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1

$egingroup$

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

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

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

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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