Question

if (1 + Cos A) (1 - cos A) = 3 / 4 find the value of secA

Answer 1

Solution:-

If you have any problem, you look in attachment.


Here, p = Perpendicular, b = Base and h = Hypotenuse

(1 + Cos A) (1 - cos A) = 3 / 4

$= > 1 - {cos}^{2} = \frac{3}{4}$

$= > {sin}^{2} = \frac{3}{4}$

$= > sinA = \frac{ \sqrt{3} }{2} = \frac{p}{h}$

$BC = \sqrt{4 - 3} = \sqrt{1} = 1$

(By Pythagoras Theorem)

$Therefore, \: secA = \frac{h}{b} = \frac{2}{1} = 2$

Answer 2

$(1 + cosa) (1 - cosa) = \frac{3}{4} \\ 1 - { \cos(a) }^{2} = \frac{3}{4} \\ - \frac{1}{ { \sec(a) }^{2} } = \frac{3 + 4}{4} \\ \frac{1}{{ \sec(a) }^{2} } = - \frac{7}{4} \\ { \sec(a ) }^{2} = \frac{ - 4}{7} \\ \sec(a ) = \sqrt{ \frac{ - 4}{7} }$

I hope this will help you