Question

1+cosA/sinA+sinA/1+cosA=2/sinA​

Answer 1

TO PROVE :–

$\\ \implies \bf \dfrac{1+ \cos A}{ \sin A}+ \dfrac{ \sin A}{1+ \cos A}= \dfrac{2}{ \sin A}$

SOLUTION :–

• Let's take L.H.S. –

$\\ \: \: \: = \: \: \: \bf \dfrac{1+ \cos A}{ \sin A}+ \dfrac{ \sin A}{1+ \cos A}$

$\\ \: \: \: = \: \: \: \bf \dfrac{(1+ \cos A)^{2} +(\sin A)^{2} }{( \sin A)(1+ \cos A)}$

$\\ \: \: \: = \: \: \: \bf \dfrac{1 +\cos^{2} A + 2\cos A+\sin^{2} A}{( \sin A)(1+ \cos A)}$

$\\ \: \: \: = \: \: \: \bf \dfrac{1 +2\cos A+ \{\sin^{2} A +\cos^{2} A \}}{( \sin A)(1+ \cos A)}$

• We know that –

$\\ \longrightarrow\bf \sin^{2} \theta+\cos^{2} \theta = 1$

• So that –

$\\ \: \: \: = \: \: \: \bf \dfrac{1 +2\cos A+ \{1\}}{( \sin A)(1+ \cos A)}$

$\\ \: \: \: = \: \: \: \bf \dfrac{1 +2\cos A+1}{( \sin A)(1+ \cos A)}$

$\\ \: \: \: = \: \: \: \bf \dfrac{2 +2\cos A}{( \sin A)(1+ \cos A)}$

$\\ \: \: \: = \: \: \: \bf \dfrac{2 \cancel{(1 +\cos A)}}{( \sin A) \cancel{(1+ \cos A)}}$

$\\ \: \: \: = \: \: \: \bf \dfrac{2}{\sin A}$

$\\ \: \: \: = \: \: \: \bf R.H.S.$

$\\ \implies\large { \boxed{ \boxed{ \bf \green{Hence \: \: Proved}}}}$

Answer 2

Step-by-step explanation:

$\large\bf{\underline{\underline{To\ prove:-}}}$

$\bf : \implies {\dfrac{1\ +\ cos\ A}{sin\ A}\ +\ \dfrac{sin\ A}{1\ +\ cos\ A}\ =\ \dfrac{2}{sin\ A}}$

$\large\bf{\underline{\underline{Finding\ solution:-}}}$

● By taking L.H.S:-

$\bf : \implies {\dfrac{1\ +\ cos\ A}{sin\ A}\ +\ \dfrac{sin\ A}{1\ +\ cos\ A}} \\ \\ \\ \bf : \implies {\dfrac{(1\ +\ cos\ A)^2\ +\ (sin\ A)^2}{(sin\ A)(1\ +\ cos\ A)}} \\ \\ \\ \bf : \implies {\dfrac{1\ +\ cos^2\ A\ +\ 2\ cos\ A\ +\ sin^2\ A}{sin\ A\ +\ 1\ +\ cos\ A}} \\ \\ \\ \bf : \implies {\dfrac{1\ +\ 2\ cos\ A\ +\ [sin^2\ A\ +\ cos^2\ A]}{(sin\ A)(1\ +\ cos\ A)}}$

$\bf{\underline{\underline{As\ we\ know\ that:-}}}$

$\bf : \implies {sin^2\ \theta \ +\ cos^2\ \theta \ =\ 1.}$

$\large\bf{\underline{\underline{So\ here:-}}}$

$\bf : \implies {\dfrac{1\ +\ 2\ cos\ A\ +\ (1)}{(sin\ A)(1\ +\ cos\ A)}} \\ \\ \\ \bf : \implies {\dfrac{1\ +\ 2\ cos\ A\ +\ 1}{(sin\ A)(1\ +\ cos\ A)}} \\ \\ \\ \bf : \implies {\dfrac{2\ +\ 2\ cos\ A}{(sin\ A)(1\ +\ cos\ A)}} \\ \\ \\ \bf : \implies {\dfrac{2(1\ +\ cos\ A)}{(sin\ A)(1\ +\ cos\ A)}} \\ \\ \\ \bf : \implies {\purple{\underline{\boxed{\bf \dfrac{2}{sin\ A}}}}}\bigstar$

$\bf : \implies {L.H.S}$

Hence Proved ✔.