Question

cos A/(1+sin A) + (1+Sin A) /cos A=2 sec sec A​

Answer 1

Step-by-step explanation:

Answer:

$\begin{gathered} \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=2secA\end{gathered}$

Step-by-step explanation:

$\begin{gathered}LHS = \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=\frac{cos^{2}A+(1+sinA)^{2}}{(1+sinA)cosA}\\=\frac{cos^{2}A+1^{2}+sin^{2}A+2sinA}{(1+sinA)cosA}\\=\frac{(cos^{2}A+sin^{2}A)+1+2sinA}{(1+sinA)cosA}\\=\frac{1+1+2sinA}{(1+sinA)cosA}\end{gathered}$

By Trigonometric identity:

• cos² A+ sin² A = 1

$\begin{gathered}=\frac{2+2sinA}{(1+sinA)cosA}\\=\frac{2(1+sinA)}{(1+sinA)cosA}\\\end{gathered}$

After cancellation,we get

$\begin{gathered}= \frac{2}{cosA}\\=2secA\\=RHS \end{gathered}$

Therefore,

$\begin{gathered} \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=2secA\end{gathered}$

Answer 2

$\large{\sf{\underline{\underline\red{Solution!!}}}}$

$:\implies\sf{\dfrac{Cos\: A}{1+Sin\:A}+\dfrac{1+Sin\;A}{Cos\:A} }$

$:\implies\sf{\dfrac{Cos^2A+[1+Sin\:A]^2}{Cos\:A[1+Sin\:A]} }$

$:\implies\sf{\dfrac{Cos^2A+1\:Sin\:^2 A+2\:Sin\:A}{Cos\:A+[1+Sin\:A]} }$

$:\implies\sf{\dfrac{2+2\:Sin\:A}{Cos\:A[1+Sin\:A]} }$

$:\implies\sf{\dfrac{2}{Cos\:A} }$

$:\implies\sf{2\:Sec\:A}$

@MissValiant❤࿐