Question

(1 +tan 0 + sec 0) (1 + coto - cosec 0)​

Answer 1

(1+tanA+secA) (1+cotA-cosecA)

$= (1+\dfrac{sinA}{cosA}+\dfrac{1}{cosA}) (1+\dfrac{cosA}{sinA}-\dfrac{1}{sinA}$

$= [\dfrac{(cosA+sinA)+1}{cosA}] [\dfrac{(cosA+sinA)-1}{sinA}]$

$= \dfrac{({cosA+sinA})^{2}-1}{cosA \times sinA}$

$= (\dfrac{{cos}^{2}A+{sin}^{2}A+2 cosA \times sinA-1}{cosA \times sinA} )$

$= \dfrac{1 + 2 cosA \times sinA -1}{cosA \times sinA}$

= 2

Answer 2

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