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Math  /  Trigonometry

Question4) Prove Trigonometric Identities (10 marks, 5 each) tan⁡θ+cot⁡θ=sec⁡θcsc⁡θcan⁡2α1+can⁡2α=sin⁡2α\tan \theta+\cot \theta=\sec \theta \csc \theta \quad \frac{\operatorname{can}^{2} \alpha}{1+\operatorname{can}^{2} \alpha}=\sin ^{2} \alpha

Studdy Solution

STEP 1

1. We are given two trigonometric identities to prove.
2. We will use fundamental trigonometric identities to simplify and prove the given identities.

STEP 2

1. Prove the identity tan⁡θ+cot⁡θ=sec⁡θcsc⁡θ\tan \theta + \cot \theta = \sec \theta \csc \theta.
2. Prove the identity can⁡2α1+can⁡2α=sin⁡2α\frac{\operatorname{can}^{2} \alpha}{1+\operatorname{can}^{2} \alpha}=\sin ^{2} \alpha.

STEP 3

Start by expressing tan⁡θ\tan \theta and cot⁡θ\cot \theta in terms of sine and cosine:
tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}

STEP 4

Add tan⁡θ\tan \theta and cot⁡θ\cot \theta using a common denominator:
tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}

STEP 5

Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1:
tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ\tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta}

STEP 6

Express sec⁡θ\sec \theta and csc⁡θ\csc \theta in terms of sine and cosine:
sec⁡θ=1cos⁡θ,csc⁡θ=1sin⁡θ\sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta}

STEP 7

Multiply sec⁡θ\sec \theta and csc⁡θ\csc \theta:
sec⁡θcsc⁡θ=1cos⁡θ⋅1sin⁡θ=1sin⁡θcos⁡θ\sec \theta \csc \theta = \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{1}{\sin \theta \cos \theta}

STEP 8

Since both sides of the equation are equal, the identity is proved:
tan⁡θ+cot⁡θ=sec⁡θcsc⁡θ\tan \theta + \cot \theta = \sec \theta \csc \theta

STEP 9

Rewrite can⁡2α\operatorname{can}^{2} \alpha as cot⁡2α\cot^2 \alpha and express it in terms of sine and cosine:
cot⁡2α=cos⁡2αsin⁡2α\cot^2 \alpha = \frac{\cos^2 \alpha}{\sin^2 \alpha}

STEP 10

Substitute cot⁡2α\cot^2 \alpha into the given identity:
cot⁡2α1+cot⁡2α=cos⁡2αsin⁡2α1+cos⁡2αsin⁡2α\frac{\cot^2 \alpha}{1 + \cot^2 \alpha} = \frac{\frac{\cos^2 \alpha}{\sin^2 \alpha}}{1 + \frac{\cos^2 \alpha}{\sin^2 \alpha}}

STEP 11

Simplify the expression by finding a common denominator:
cos⁡2αsin⁡2αsin⁡2α+cos⁡2αsin⁡2α=cos⁡2αsin⁡2α+cos⁡2α\frac{\frac{\cos^2 \alpha}{\sin^2 \alpha}}{\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin^2 \alpha}} = \frac{\cos^2 \alpha}{\sin^2 \alpha + \cos^2 \alpha}

STEP 12

Use the Pythagorean identity sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1:
cos⁡2α1=cos⁡2α\frac{\cos^2 \alpha}{1} = \cos^2 \alpha

STEP 13

Recognize that cos⁡2α=1−sin⁡2α\cos^2 \alpha = 1 - \sin^2 \alpha and rearrange:
1−cos⁡2α=sin⁡2α1 - \cos^2 \alpha = \sin^2 \alpha
Both identities have been proved.

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