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Mathematics

Prove that:

(1+sin θ)2+(1−sin θ)22 cos 2θ\dfrac{(1 + \text{sin }θ)^2 + (1 - \text{sin }θ)^2}{2\text{ cos }^2θ} = sec2θ + tan2θ

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Answer

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

⇒(1+sin θ)2+(1−sin θ)22cos 2θ⇒1+sin 2θ+2sin θ+1+sin 2θ−2sin θ2cos 2θ⇒2+2sin 2θ2cos 2θ⇒2(1+sin 2θ)2cos 2θ⇒1+sin 2θcos 2θ⇒1cos 2θ+sin 2θcos 2θ⇒sec θ2+tan 2θ\Rightarrow \dfrac{(1 + \text{sin }θ)^2 + (1 - \text{sin }θ)^2}{2\text{cos }^2θ} \\[1em] \Rightarrow\dfrac{1 + \text{sin }^2θ + 2\text{sin }θ + 1 + \text{sin }^2θ - 2\text{sin }θ}{2\text{cos }^2θ}\\[1em] \Rightarrow\dfrac{2 + 2\text{sin }^2θ}{2\text{cos }^2θ}\\[1em] \Rightarrow\dfrac{2(1 + \text{sin }^2θ)}{2\text{cos }^2θ}\\[1em] \Rightarrow\dfrac{1 + \text{sin }^2θ}{\text{cos }^2θ}\\[1em] \Rightarrow\dfrac{1}{\text{cos }^2θ} + \dfrac{\text{sin }^2θ}{\text{cos }^2θ}\\[1em] \Rightarrow \text{sec }θ^2 + \text{tan }^2θ\\[1em]

Since, L.H.S. = sec2θ + tan2θ = R.H.S.

Hence, proved (1+sin θ)2+(1−sin θ)22cos 2θ\dfrac{(1 + \text{sin }θ)^2 + (1 - \text{sin }θ)^2}{2\text{cos }^2θ} = sec2θ + tan2θ.

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