<s>[INST] To find the Value of $\tan A + \cot A$, if the value of $\sin A + \cos A$ is given 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


*

*$\sin^2 A + \cos^2 A = 1$

*$1 + \cot^2 A = \csc^2 A$

*$\tan^2A + 1 = \sec^2 A$

 [/INST] $\cos\frac {\pi}4=\sin\frac {\pi}4=\frac1{\sqrt{2}}$. 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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