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Mathematics

Prove the following identity, where the angles involved are acute angles for which the trigonometric ratios as defined:

(1−tan θ1−cot θ)2\Big(\dfrac{1 - \text{tan } θ}{1 - \text{cot } θ}\Big)^2 = tan2 θ

Trigonometric Identities

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Answer

The L.H.S of above equation can be written as,

⇒(1−tan θ1−cot θ)2⇒(1−tan θ1−1tan θ)2⇒((1−tan θ).tan θtan θ−1)2⇒(−(tan θ−1).tan θtan θ−1)2⇒(−tan θ)2⇒tan 2θ.\Rightarrow \Big(\dfrac{1 - \text{tan } θ}{1 - \text{cot } θ}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1 - \text{tan } θ}{1 - \dfrac{1}{\text{tan } θ}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{(1 - \text{tan } θ). \text{tan } θ}{\text{tan } θ - 1} \Big)^2 \\[1em] \Rightarrow \Big(\dfrac{-(\text{tan } θ - 1). \text{tan } θ}{\text{tan } θ - 1} \Big)^2 \\[1em] \Rightarrow (-\text{tan } θ)^2 \\[1em] \Rightarrow \text{tan }^2 θ.

Since, L.H.S. = R.H.S.

Hence, proved (1−tan θ1−cot θ)2\Big(\dfrac{1 - \text{tan } θ}{1 - \text{cot } θ}\Big)^2 = tan2 θ

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