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

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Final answer to the problem

true

Step-by-step Solution

How should I solve this problem?

  • Dimostrare da RHS (lato destro)
  • Dimostrare dal LHS (lato sinistro)
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  • Equazione differenziale esatta
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Starting from the right-hand side (RHS) of the identity

$\frac{\cos\left(x\right)}{1+\sin\left(x\right)}$

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$\frac{\cos\left(x\right)}{1+\sin\left(x\right)}$

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Learn how to solve problems step by step online. (1-sin(x))/cos(x)=cos(x)/(1+sin(x)). Starting from the right-hand side (RHS) of the identity. Apply the formula: \frac{a}{b}=\frac{a}{b}\frac{conjugate\left(b\right)}{conjugate\left(b\right)}, where a=\cos\left(x\right), b=1+\sin\left(x\right) and a/b=\frac{\cos\left(x\right)}{1+\sin\left(x\right)}. Apply the formula: \frac{a}{b}\frac{c}{f}=\frac{ac}{bf}, where a=\cos\left(x\right), b=1+\sin\left(x\right), c=1-\sin\left(x\right), a/b=\frac{\cos\left(x\right)}{1+\sin\left(x\right)}, f=1-\sin\left(x\right), c/f=\frac{1-\sin\left(x\right)}{1-\sin\left(x\right)} and a/bc/f=\frac{\cos\left(x\right)}{1+\sin\left(x\right)}\frac{1-\sin\left(x\right)}{1-\sin\left(x\right)}. Apply the formula: \left(a+b\right)\left(a+c\right)=a^2-b^2, where a=1, b=\sin\left(x\right), c=-\sin\left(x\right), a+c=1-\sin\left(x\right) and a+b=1+\sin\left(x\right).

Final answer to the problem

true

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