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Math  /  Trigonometry

QuestionProve the identity (sec⁡θ−cos⁡θ)(csc⁡θ−sin⁡θ)=tan⁡θ1+tan⁡2θ(\sec \theta-\cos \theta)(\csc \theta-\sin \theta)=\frac{\tan \theta}{1+\tan ^{2} \theta}

Studdy Solution

STEP 1

1. The identity involves trigonometric functions: secant, cosine, cosecant, sine, and tangent.
2. We will use trigonometric identities to simplify and prove the given identity.

STEP 2

1. Simplify the left-hand side (LHS) of the identity.
2. Simplify the right-hand side (RHS) of the identity.
3. Show that the simplified LHS equals the simplified RHS.

STEP 3

Start by expressing sec⁡θ\sec \theta and csc⁡θ\csc \theta in terms of cos⁡θ\cos \theta and sin⁡θ\sin \theta:
sec⁡θ=1cos⁡θ,csc⁡θ=1sin⁡θ\sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta}
Substitute these into the LHS:
(sec⁡θ−cos⁡θ)(csc⁡θ−sin⁡θ)=(1cos⁡θ−cos⁡θ)(1sin⁡θ−sin⁡θ)(\sec \theta - \cos \theta)(\csc \theta - \sin \theta) = \left(\frac{1}{\cos \theta} - \cos \theta\right)\left(\frac{1}{\sin \theta} - \sin \theta\right)

STEP 4

Simplify the expression inside each parenthesis:
1cos⁡θ−cos⁡θ=1−cos⁡2θcos⁡θ=sin⁡2θcos⁡θ\frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta}
1sin⁡θ−sin⁡θ=1−sin⁡2θsin⁡θ=cos⁡2θsin⁡θ\frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta}

STEP 5

Multiply the simplified expressions:
\left(\frac{\sin^2 \theta}{\cos \theta}\right)\left(\frac{\cos^2 \theta}{\sin \theta}\right) = \frac{\sin^2 \theta \cdot \cos^2 \theta}{\cos \theta \cdot \sin \theta} = \frac{\sin \theta \cdot \cos \theta}

STEP 6

Simplify the RHS of the identity:
tan⁡θ1+tan⁡2θ\frac{\tan \theta}{1 + \tan^2 \theta}
Recall the identity 1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta, so:
tan⁡θsec⁡2θ=sin⁡θ/cos⁡θ1/cos⁡2θ=sin⁡θ⋅cos⁡θ\frac{\tan \theta}{\sec^2 \theta} = \frac{\sin \theta / \cos \theta}{1 / \cos^2 \theta} = \sin \theta \cdot \cos \theta

STEP 7

Compare the simplified LHS and RHS:
Both simplify to sin⁡θ⋅cos⁡θ\sin \theta \cdot \cos \theta, thus proving the identity:
(sec⁡θ−cos⁡θ)(csc⁡θ−sin⁡θ)=tan⁡θ1+tan⁡2θ(\sec \theta - \cos \theta)(\csc \theta - \sin \theta) = \frac{\tan \theta}{1 + \tan^2 \theta}
The identity is proven.

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