Pls answer the question step by step
Answers
Step-by-step explanation:
This is not 5 points question, because this is easy but yeah not that easy,
first of all lets do few things in advance
(1)
\begin{lgathered}\frac{1}{sin^{2}x} + \frac{1}{cos^{2}x} \\\\= \frac{sin^{2}x+cos^{2}x}{sin^{2}xcos^{2}x} \\ \\ = \frac{1}{sin^{2}xcos^{2}x}\end{lgathered}
sin
2
x
1
+
cos
2
x
1
=
sin
2
xcos
2
x
sin
2
x+cos
2
x
=
sin
2
xcos
2
x
1
(2)
\begin{lgathered}\frac{sinx}{cosx} + \frac{cosx}{sinx} \\ \\ = \frac{sin^{2}x + cos^{2}x}{sinxcosx} \\ \\ \frac{1}{sinxcosx}\end{lgathered}
cosx
sinx
+
sinx
cosx
=
sinxcosx
sin
2
x+cos
2
x
sinxcosx
1
Proof;
\begin{lgathered}(sinA + \frac{1}{cosA})^{2} + (cosA + \frac{1}{sinA})^{2} \\ \\ sin^{2}A + \frac{1}{cos^{2}A} + 2 \frac{sinA}{cosA} + cos^{2}A + \frac{1}{sin^{2}A} + 2 \frac{cosA}{sinA} \\ \\\end{lgathered}
(sinA+
cosA
1
)
2
+(cosA+
sinA
1
)
2
sin
2
A+
cos
2
A
1
+2
cosA
sinA
+cos
2
A+
sin
2
A
1
+2
sinA
cosA
\begin{lgathered}[sin^{2}A + cos^{2}A] + [\frac{1}{sin^{2}A} + \frac{1}{cos^{2}A}] + 2[\frac{sinA}{cosA} + \frac{cosA}{sinA}] \\ \\\end{lgathered}
[sin
2
A+cos
2
A]+[
sin
2
A
1
+
cos
2
A
1
]+2[
cosA
sinA
+
sinA
cosA
]
\begin{lgathered}1 + \frac{1}{sin^{2}Acos^{2}A} + \frac{2}{sinAcosA} \\ \\ (1 + \frac{1}{sinAcosA})^{2} \\ \\ (1 + secAcosecA)^{2}\end{lgathered}
1+
sin
2
Acos
2
A
1
+
sinAcosA
2
(1+
sinAcosA
1
)
2
(1+secAcosecA)
2