(1+tan 0 + sec 0) (1 + coto - cosec 0)
Answers
Answer:
(1+tanA+secA) (1+cotA-cosecA)
\begin{gathered}\\ \\\end{gathered}
= (1+\dfrac{sinA}{cosA}+\dfrac{1}{cosA}) (1+\dfrac{cosA}{sinA}-\dfrac{1}{sinA}=(1+
cosA
sinA
+
cosA
1
)(1+
sinA
cosA
−
sinA
1
\begin{gathered}\\ \\\end{gathered}
= [\dfrac{(cosA+sinA)+1}{cosA}] [\dfrac{(cosA+sinA)-1}{sinA}]=[
cosA
(cosA+sinA)+1
][
sinA
(cosA+sinA)−1
]
\begin{gathered}\\ \\\end{gathered}
= \dfrac{({cosA+sinA})^{2}-1}{cosA \times sinA}=
cosA×sinA
(cosA+sinA)
2
−1
\begin{gathered}\\ \\\end{gathered}
= (\dfrac{{cos}^{2}A+{sin}^{2}A+2 cosA \times sinA-1}{cosA \times sinA} )=(
cosA×sinA
cos
2
A+sin
2
A+2cosA×sinA−1
)
\begin{gathered}\\ \\\end{gathered}
= \dfrac{1 + 2 cosA \times sinA -1}{cosA \times sinA}=
cosA×sinA
1+2cosA×sinA−1
\begin{gathered}\\ \\\end{gathered}
= 2