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5 votes
Prove
cos A /(1- sin A) = (1 + sin A)/cos A​

User Drlobo
by
8.2k points

1 Answer

5 votes

Answer:

answer is in exaplation

Explanation:

cosA

+

cosA

1+sinA

=2secA

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

User Fordareh
by
8.3k points

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