Question

Demuestra las siguientes identidades trigonométricas. ​

Answer 1

Respuesta:

Explicación paso a paso:

Tema: $\large\textsf{Identidades trigonom\'etricas}$

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1) $\large\dfrac{\text{Cosx}\times\text{Secx}}{\text{Tanx}} =\text{Ctgx}$

• $\boxed{\bold{Cosx\times Secx=1}}$

$\large\dfrac{1}{\text{Tanx}} =\text{Ctgx}$

• $\boxed{\bold{\dfrac{1}{Tanx} =Ctgx}}$

$\huge\boxed{\bold{Ctgx=Ctgx}}$

3) $\large(1-\text{Senx})(1+\text{Senx})=\text{Cos}^{2} x$

• $\bold{(a-b)(a+b)=a^{2} -b^{2}}$

$\large1^{2} -\text{Sen}^{2}x=\text{Cos}^{2}x$

• $\boxed{\bold{Cos^{2}x=1-Sen^{2}x}}$

$\huge\boxed{\bold{Cos^{2}x=Cos^{2}x}}$

4) $\large(1-\text{Cosx})(1+\text{Cosx})=\dfrac{1}{\text{Csc}^{2}x}$

• $\bold{(a-b)(a+b)=a^{2} -b^{2}}$

$\large1^{2} -\text{Cos}^{2} =\dfrac{1}{\text{Csc}^{2}x}$

• $\boxed{\bold{Sen^{2}x=1-Cos^{2}x}}$

$\large\text{Sen}^{2}x=\dfrac{1}{\text{Csc}^{2}x}$

• $\boxed{\bold{\dfrac{1}{Cscx^{2}x} =Sen^{2}x}}$

$\huge\boxed{\bold{Sen^{2} x=Sen^{2}x}}$

5) $\large\dfrac{\text{Cosx}}{\text{Secx}}+ \dfrac{\text{Senx}}{\text{Cscx}} =1$

• $\boxed{\bold{Secx=\dfrac{1}{Cosx}}}$
• $\boxed{\bold{Cscx=\dfrac{1}{Senx}}}$

$\large\dfrac{\text{Cosx}}{\dfrac{1}{\text{Cosx}}} +\dfrac{\text{Senx}}{\dfrac{1}{\text{Senx}}} =1$

$\large\dfrac{\text{Cos}^{2}x}{1} +\dfrac{\text{Sen}^{2}x}{1} =1$

• $\boxed{\bold{Cos^{2}x+Sen^{2}x=1}}$

$\large\text{Cos}^{2} x+\text{Sen}^{2} x=1$

$\huge\boxed{\bold{1=1}}$

6) $\large\text{Sen}^{4}x-\text{Cos}^{4}x=\text{Sen}^{2}x-\text{Cos}^{2}x$

• $\bold{(a-b)(a+b)=a^{2} -b^{2}}$

$\large(\text{Sen}^{2}x+\text{Cos}^{2}x)(\text{Sen}^{2}-\text{Cos}^{2} )=\text{Sen}^{2}x-\text{Cos}^{2}x$

• $\boxed{\bold{Cos^{2}x+Sen^{2}x=1}}$

$\large(1)(\text{Sen}^{2}-\text{Cos}^{2} )=\text{Sen}^{2}x-\text{Cos}^{2}x$

$\huge\boxed{\bold{Sen^{2}-Cos^{2} =Sen^{2}x-Cos^{2}x}}$

[Las demás que faltan las estaré adjuntando, saludos]