18. Prove that
COSA
COSA
1 + sinA 1-sin A
2 sec A
Answers
Answer:
cosA
+
cosA
1+sinA
=2secA
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}
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{gathered}=\frac{2+2sinA}{(1+sinA)cosA}\\=\frac{2(1+sinA)}{(1+sinA)cosA}\\\end{gathered}
=
(1+sinA)cosA
2+2sinA
=
(1+sinA)cosA
2(1+sinA)
After cancellation,we get
\begin{gathered}= \frac{2}{cosA}\\=2secA\\=RHS \end{gathered}
=
cosA
2
=2secA
=RHS
Therefore,
\begin{gathered} \frac{cosA}{1+sinA}+\frac{1+sinA}{cosA}\\=2secA\end{gathered}
1+sinA
cosA
+
cosA
1+sinA
=2secA