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
- $\sin^2 A + \cos^2 A = 1$
- $1 + \cot^2 A = \csc^2 A$
- $\tan^2A + 1 = \sec^2 A$
5 Answers
$\begingroup$$$\tan{A}+\cot{A}=\frac{1}{\sin{A}\cos{A}}=\frac{2}{(\sin{A}+\cos{A})^2-1}=2$$
$\endgroup$ 19 $\begingroup$Your second attempt is actually what helps here. Note:
$$\begin{align}\tan A+\cot A&=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\\&=\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}\\&=\frac{1}{\sin A\cos A}\\&=\frac{1}{1/2}\\&=2\end{align}$$
$\endgroup$ 5 $\begingroup$$$\begin{align}\tan A+\cot A&=\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}\\&=\dfrac{\sin^2A+\cos^2A}{\sin A\cos A}\\&=\dfrac1{\sin A\cos A}\\&=\dfrac2{2\sin A\cos A}\\&=\dfrac2{(\sin A+\cos A)^2-1}\\&=\dfrac2{(\sqrt2)^2-1}\\&=2\end{align}$$
$\endgroup$ $\begingroup$$\cos\frac {\pi}4=\sin\frac {\pi}4=\frac1{\sqrt{2}}$. Hence what you have got is $$\cos\frac {\pi}4\sin A+\sin\frac {\pi}4\cos A=1$$ or $\sin\Big(A+\frac {\pi}4\Big)=1$. Can you solve from here?
$\endgroup$ 7 $\begingroup$Let's use $\sin^2A=\sqrt{1-\cos^2A}$
Equation transform into$$\sin A+\sqrt{1-\sin^2A}=\sqrt2$$
Which is quadratic equation and easy to solve
$\endgroup$ 2
