Harder Trigonometry Identity ($\sec A+\csc A$)

Question

How do I prove:

$\sin A (1 + \tan A) + \cos A (1 + \cot A) = \sec A + \csc A$

I've tried expanding the brackets by multiplying sin A and cos A to the left hand side but to no avail. Where should I start from?

Answer 1

$\tan A=\sin A/\cos A$, and $\cot A=\cos A/\sin A$, so that making this substitution, we have \begin{align} \sin A\left(1+\frac{\sin A}{\cos A}\right)+\cos A\left(1+\frac{\cos A}{\sin A}\right) & =\sin A + \frac{\sin^2 A}{\cos A}+\cos A +\frac{\cos^2 A}{\sin A} \\ & =\sin A+\cos A+\frac{1-\cos^2 A}{\cos A}+\frac{1-\sin^2 A}{\sin A} \\ & = \sin A+\cos A+\sec A-\cos A+\csc A-\sin A \\ & = \sec A +\csc A \end{align}

Where I also made use of $\cos^2 A+\sin^2 A=1$.

Answer 2

Hint: LHS= $s(1+s/c)+c(1+c/s)=\dots=\dfrac{s(s^2+c^2)+c(s^2+c^2)}{sc}$.

Answer 3

Another possibility, just continuing to get common denominators:

$$\sin A \left(1 + {\sin A\over \cos A}\right) + \cos A\left(1 + {\cos A\over \sin A}\right) \\ = \sin A \left({\cos A + \sin A\over \cos A}\right) + \cos A \left({\sin A + \cos A\over \sin A}\right) \\ = {\sin A(\cos A + \sin A)\over \cos A} + {\cos A (\sin A + \cos A)\over \sin A}\\= {\sin^2 A(\cos A + \sin A) + \cos^2 A(\sin A + \cos A)\over \cos A \sin A} \\ = {(\sin^2 A + \cos^2 A)(\cos A + \sin A)\over \cos A \sin A} \\ = {\cos A + \sin A\over \cos A \sin A} \\ = {1\over \sin A} + {1\over \cos A}$$

Answer 4

\begin{align} \sin A \frac{(\sin A+\cos A)}{\cos A}&+\cos A\frac{(\sin A+\cos A)}{\sin A} \\[6px]&= (\sin A+\cos A)\frac{\sin A}{\cos A}+(\sin A+\cos A)\frac{\cos A}{\sin A} \\[6px]&= (\cos A+\sin A)\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right) \\[6px]&= (\cos A+\sin A)\frac{(\sin^2 A+\cos^2 A)}{(\cos A\sin A)} \\[6px]&= \frac{\cos A}{\cos A\sin A}+\frac{\sin A}{\cos A\sin A} \\[6px]&= \frac1{\sin A}+\frac1{\cos A} \\[6px]&= \csc A+\sec A \end{align}

Answer 5

Starting from,

$\sin A (1 + \tan A) + \cos A (1 + \cot A)\\ \implies \sin A+\frac{\sin^2A}{\cos A}+\cos A + \frac{\cos^2 A}{\sin A}\\ \implies \sin A+ \sec A -\cos A+\cos A+\csc A -\sin A \\ \implies \sec A+\csc A=\text{R.H.S}$

Answer 6

So expand... sinA(1 + tanA) + cosA(1 + cotA) = secA+cscA

==> sinA(1 + sinA/cosA) + cosA(1 + cosA/sinA) = secA + cscA

==> sinA + sin^2A/cosA + cosA + cos^2A/sinA

==> sinA + cos^2A/sinA + sin^2A/cosA + cosA

==> (sin^2A + cos^2A)/sinA + (sin^2A + cos^2A)/cosA

==>1/sinA + 1/cosA

==> cscA + secA