Prove
cos A /(1- sin A) = (1 + sin A)/cos A

answer is in exaplation

Step-by-step explanation:

cosA

+

cosA

1+sinA

=2secA

Step-by-step explanation:

\begin{lgathered}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{lgathered}

LHS=

1+sinA

cosA

+

cosA

1+sinA

=

(1+sinA)cosA

cos

2

A+(1+sinA)

2

=

(1+sinA)cosA

cos

2

A+1

2

+sin

2

A+2sinA

=

(1+sinA)cosA

(cos

2

A+sin

2

A)+1+2sinA

=

(1+sinA)cosA

1+1+2sinA

/* By Trigonometric identity:

cos² A+ sin² A = 1 */

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

=

(1+sinA)cosA

2+2sinA

=

(1+sinA)cosA

2(1+sinA)

After cancellation,we get

\begin{lgathered}= \frac{2}{cosA}\\=2secA\\=RHS\end{lgathered}

=

cosA

2

=2secA

=RHS

Therefore,

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

1+sinA

cosA

+

cosA

1+sinA

=2secA

given hc ∥ gd.¯¯¯¯¯¯which statement is true

step-by-step explanation:

∠cbe and ∠deb are same-side interior angles