Question

if (1+sina)(1+sinb)(1+sinc)=(1-sina)(1-sinb)(1-sinc)=k then value of k=?
a. (cosacosbcosc)
b.(sinasinbsinc)
c. (3sinasinbsinc)

Answer 1

Okay so in this queston we know that the ...
First big thing is k
And even the second big thing is k right?
[Pardon my mathematical term use, I am on my iPad now]

Answer 2

The answer is a) btw.

Answer 3

consider:$$k=(1+\sin a)(1+\sin b)(1+\sin c)\\k=(1-\sin a)(1-\sin b)(1-\sin c)$$hence$$k^2=(1+\sin a)(1-\sin a)(1+\sin b)(1-\sin b)(1+\sin c)(1-\sin c)\\k^2=(1-\sin^2 a)(1-\sin^2 b)(1-\sin^2 c)\\k^2=\cos^2a\cos^2b\cos^2c$$hence \(k=\cos a\cos b\cos c\)

Answer 4

explain me

Answer 5

@Sandeepsra since \(k\) is equal to both it follows that their product is \(k^2\); if you distribute you get differences of squares that give \((1-\sin^2 a)(1-\sin^2 b)(1-\sin^2c)=\cos^2 a\cos^2b\cos^2c\) by the Pythagorean identity. we then merely take the square root to get \(k=\cos a\cos b\cos c\)

Answer 6

Yeah just multiply them and use...
a2-b2 identity and then the basic Trigo identity....
sin2theta + cos2theta = 1

Answer 7

please explain me in detail

Answer 8

@Sandeepsra $$k=(1+\sin a)(1+\sin b)(1+\sin c)\\k=(1-\sin a)(1-\sin b)(1-\sin c)$$hence$$k\cdot k=(1+\sin a)(1+\sin b)(1+\sin c)(1-\sin a)(1-\sin b)(1-\sin c)\\k^2=(1+\sin a)(1-\sin a)(1+\sin b)(1-\sin b)(1+\sin c)(1-\sin c)$$recall that \((a+b)(a-b)=a^2-b^2\) hence \((1+\sin u)(1-\sin u)=1-\sin^2 u\):$$k^2=(1-\sin^2a)(1-\sin^2b)(1-\sin^2c)$$now apply the Pythagorean identity; remember \(\cos^2u+\sin^2u=1\) hence \(\cos^2u=1-\sin^2u\) giving us:$$k^2=\cos^2 a\cos^2 b\cos^2 c$$take the square root of both sides:$$\sqrt{k^2}=\sqrt{\cos^2a\cos^2b\cos^2c}\\k=\cos a\cos b\cos c$$

Answer 9

ok