Prove that 1 + (tan^2 θ) / (1 + sec θ) = sec θ

The L.H.S of above equation can be written as, Since, L.H.S. = R.H.S. Hence, proved 1 + = sec θ .

Mathematics

Prove the following identity, where the angles involved are acute angles for which the trigonometric ratios as defined:
1 + $\dfrac{\text{tan }^2 θ}{1 + \text{sec } θ}$ = sec θ

Trigonometric Identities

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Answer

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

$\Rightarrow 1 + \dfrac{\text{tan }^2 θ}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \dfrac{\text{sec }^2 θ - 1}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \dfrac{(\text{sec } θ - 1)(\text{sec } θ + 1)}{1 + \text{sec } θ} \\[1em] \Rightarrow 1 + \text{sec } θ - 1\\[1em] \Rightarrow \text{sec } θ.$

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

Hence, proved 1 + $\dfrac{\text{tan }^2 θ}{1 + \text{sec } θ}$ = sec θ .

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Mathematics

Prove the following identity, where the angles involved are acute angles for which the trigonometric ratios as defined:
1 + $\dfrac{\text{tan }^2 θ}{1 + \text{sec } θ}$ = sec θ

Trigonometric Identities

Answer

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