How do you prove $(\frac{{sin}^{2} x + 2 cos x - 1}{2 + cos x - {cos}^{2} x}) = \frac{1}{1 + sec x}$ $((sin^2x + 2cosx - 1)/(2 + cosx-cos^2x)) = 1/(1 + secx)$ ?

Question

5930 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-17T20:34:17

Left side Numerator: $(1 - {cos}^{2} x + 2 cos x - 1) = - cos x (cos x - 2)$ $(1 - cos^2 x + 2cos x -1) = -cos x(cos x - 2)$

Left side Denominator: $- (cos x + 1) (cos x - 2) .$ $-(cos x + 1)(cos x - 2).$

Left side after simplification: $\frac{cos x}{cos x + 1}$ $cos x/(cos x + 1)$

Right side: $\frac{1}{1 + \frac{1}{cos x}} = \frac{cos x}{cos x + 1}$ $1/(1 + 1/cos x) = cos x/(cos x + 1)$

Finally, left side = right side and the relation is proven.

How do you prove (sin2x+2cosx−12+cosx−cos2x)=11+secx((sin^2x + 2cosx - 1)/(2 + cosx-cos^2x)) = 1/(1 + secx)?