Question

1+sinθ÷cosθ+cosθ÷1+sinθ=2sec Plzzz Help Me​

Answer 1

Hello !

Given:-

• L.H.S = $\dfrac{1+sinA}{cosA}$ + $\dfrac{cosA}{1+sinA}$

• R.H.S = 2secA

Solution:-

L.H.S =

$\implies$ $\dfrac{1+sinA}{cosA}$ + $\dfrac{cosA}{1+sinA}$

$\implies$ $\dfrac{(1+SinA)²+cos²A}{CosA(1+sinA)}$

$\implies$ $\dfrac{1+sin²A+2SinA+cos²A}{(1+sinA)cosA}$

$\implies$ $\dfrac{1+1+2SinA}{(1+sinA)CosA}$

$\implies$ $\dfrac{2+2sinA}{(1+sinA)cosA}$

$\implies$ $\dfrac{2(1+SinA)}{(1+sinA)cosA}$

$\implies$ $\cancel\dfrac{2(1+sinA)}{(1+sinA)cosA}$

$\implies$ $\dfrac{2}{cosA}$ = 2secA = R.H.S

since, L.H.S = R.H.S

hence verified.

I hope it will be helpful dear ☺️❤️