Question

prove that -
1+sinØ/cosØ +cosØ/1+sinØ = 2secØ

​

Answer 1

Step-by-step explanation:

1+sin/cos + cos/1+sin

(1+sin)^2 + cos^2 / cos (1+sin)

1 + 2sin + sin^2 + cos^2 / cos (1+sin)

1 + 1 + 2sin / cos (1+sin)....{sin^2 + cos^2 = 1}

2+2sin / cos (1+sin)

2 (1+sin) / cos (1+sin)

after cancelling (1+sin) above and below,

2/ cos = 2* 1/ cos

that is 2 sec

hence proved ..

Answer 2

L.H.S :-

$\longrightarrow$ $\sf\dfrac{1+SinØ}{CosØ}$ + $\sf\dfrac{CosØ}{1+SinØ}$

$\longrightarrow$ $\sf\dfrac{(1+SinØ)²+Cos²Ø}{CosØ(1+SinØ)}$

$\longrightarrow$ $\sf\dfrac{1+Sin²Ø+2SinØ+Cos²Ø}{CosØ(1+SinØ)}$

$\longrightarrow$ $\sf\dfrac{1+1+2SinØ}{CosØ(1+SinØ)}$

$\longrightarrow$ $\sf\dfrac{2+2SinØ}{CosØ(1+SinØ)}$

$\longrightarrow$ $\sf\cancel\dfrac{2(1+SinØ)}{(1+SinØ)CosØ}$

$\longrightarrow$ $\sf\dfrac{2}{CosØ}$ = 2SecØ = R.H.S