Question

identidad trigonométricas​

Answer 1

Explicación paso a pao:

yo tambien la necesito plissss

Answer 2

Respuesta:

Explicación paso a paso:

$(a) \left csc(x)tan(x)=sec(x)$

$\\csc(x)tan(x)=sec(x)\\\frac{1}{sin(x)}\frac{sin(x)}{cos(x)} = sec(x)\\\frac{1}{cos(x)}=sec(x)\\sec(x)=sec(x)$

$(b)\left2sin^2(x)+cos^2(x)=1+sin^2(x)$

$sin^2(x)+sin^2(x)+cos^2(x)=1+sin^2(x)\\sin^2(x)+1=1+sin^2(x)\\1+sin^2(x)=1+sin^2(x)$

$(c)\left\frac{1-cos(x)}{1+cos(x)}=(csc(x)-cot(x))^2$

$\frac{1-cos(x)}{1+cos(x)}=(csc(x)-cot(x))^2\\\frac{1-cos(x)}{1+cos(x)}\frac{1-cos(x)}{1-cos(x)} = (csc(x)-cot(x))^2\\\frac{(1-cos(x))^2}{1-cos^2(x)} = (csc(x)-cot(x))^2\\\frac{(1-cos(x))^2}{sin^2(x)} = (csc(x)-cot(x))^2\\\left(\frac{1-cos(x)}{sin(x)}\right)^2 = (csc(x)-cot(x))^2\\\left(\frac{1}{sin(x)}-\frac{cos(x)}{sin(x)} \right) ^2 = (csc(x)-cot(x))^2\\(csc(x)-cot(x))^2=(csc(x)-cot(x))^2$

$(d)\frac{1-sin(x)}{(sec(x)-tan(x))^2} = 1+sin(x)$

$\frac{1-sin(x)}{(sec(x)-tan(x))^2} = 1+sin(x)\\\frac{1-sin(x)}{\left(\frac{1}{cos(x)} -\frac{sin(x)}{cos(x)} \right)^2} = 1+sin(x)\\\frac{1-sin(x)}{\left(\frac{1-sin(x)}{cos(x)}\right)^2 } = 1+sin(x)\\\frac{1-sin(x)}{\frac{(1-sin(x))^2}{(cos(x))^2} } = 1+sin(x)\\\frac{(1-sin(x))(cos^2(x))}{(1-sin(x))^2} = 1+sin(x)\\\frac{(cos^2(x))}{(1-sin(x))} = 1+sin(x)\\\frac{1-sin^2(x)}{1-sin(x)} = 1+sin(x)\\\frac{(1+sin(x))(1-sin(x))}{1-sin(x)} = 1+sin(x)\\1+sin(x)=1+sin(x)$