# Step-by-step Solution

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## Step-by-step Solution

Problem to solve:

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

Choose the solving method

1

Multiply and divide the fraction $\frac{\cos\left(x\right)}{1+\sin\left(x\right)}$ by the conjugate of it's denominator $1+\sin\left(x\right)$

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

Learn how to solve trigonometric identities problems step by step online.

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

Learn how to solve trigonometric identities problems step by step online. Prove the trigonometric identity (1-sin(x))/(cos(x)=(cos(x)/(1+sin(x)). Multiply and divide the fraction \frac{\cos\left(x\right)}{1+\sin\left(x\right)} by the conjugate of it's denominator 1+\sin\left(x\right). Multiplying fractions \frac{\cos\left(x\right)}{1+\sin\left(x\right)} \times \frac{1-\sin\left(x\right)}{1-\sin\left(x\right)}. The sum of two terms multiplied by their difference is equal to the square of the first term minus the square of the second term. In other words: (a+b)(a-b)=a^2-b^2, where:<ul><li>The first term (a) is 1.</li><li>The second term (b) is \sin\left(x\right).</li></ul>Then:. Apply the trigonometric identity: 1-\sin\left(x\right)^2=\cos\left(x\right)^2.

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

### Main topic:

Trigonometric Identities

~ 0.14 s