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Question

Mathematics Question on Trigonometric Functions

Given that p=tan⁡ α+tan⁡ β,p = \tan \, \alpha+ \tan \, \beta, and q=cot⁡ α+cot⁡ βq = \cot \, \alpha + \cot \, \beta ; then what is (1p−1q)\left( \frac{1}{p} - \frac{1}{q} \right) equal to ?

A

cot⁡(α−β)\cot (\alpha - \beta)

B

tan⁡(α−β)\tan (\alpha - \beta)

C

tan⁡(α+β)\tan (\alpha + \beta)

D

cot⁡(α+β)\cot (\alpha + \beta)

Answer

cot⁡(α+β)\cot (\alpha + \beta)

Explanation

Solution

Since, p=tan⁡ α+tan⁡ βp = \tan \, \alpha + \tan \, \beta and q=cot⁡ α+cot⁡ βq = \cot \, \alpha + \cot \, \beta q=tan⁡ α+tan⁡ βq = \tan \, \alpha + \tan \, \beta ⇒q=1tan⁡α+1tan⁡β=tan⁡α+tan⁡βtan⁡αtan⁡β\Rightarrow q = \frac{1}{\tan\alpha} + \frac{1}{\tan\beta} = \frac{\tan\alpha + \tan\beta}{\tan\alpha \tan\beta} q=ptan⁡αtan⁡βq = \frac{p}{\tan\alpha \tan\beta} Hence , 1p−1q=1p−tan⁡αtan⁡βp\frac{1}{p} - \frac{1}{q} = \frac{1}{p} - \frac{\tan\alpha \tan\beta}{p} =1−tan⁡αtan⁡βp= \frac{1- \tan\alpha \tan\beta}{p} =1−tan⁡αtan⁡βtan⁡α+tan⁡β=1tan⁡(α+β)= \frac{1- \tan\alpha \tan\beta}{\tan\alpha + \tan\beta} = \frac{1}{\tan\left(\alpha+\beta\right)} =cot⁡(α+β)= \cot \left(\alpha+\beta\right)