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
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 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
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
0
\$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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\$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
–user427802
Dec 31, 2017 at 4:44
0
\$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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