Question : Which of the following is equal to $[\frac{ an heta+\sec heta–1}{ an heta–\sec heta+1}]$?
Option 1: $\frac{1+\sin heta}{\cos heta}$
Option 2: $\frac{1+ an heta}{\cot heta}$
Option 3: $\frac{1+\cot heta}{ an heta}$
Option 4: $\frac{1+\cos heta}{\sin heta}$

Question

Question : Which of the following is equal to $[\frac{ an heta+\sec heta–1}{ an heta–\sec heta+1}]$?

Option 1: $\frac{1+\sin heta}{\cos heta}$

Option 2: $\frac{1+ an heta}{\cot heta}$

Option 3: $\frac{1+\cot heta}{ an heta}$

Option 4: $\frac{1+\cos heta}{\sin heta}$

Team Careers360 · Accepted Answer

Correct Answer: $\frac{1+\sin heta}{\cos heta}$

Solution : Given: The given trigonometric expression is $[\frac{ an heta+\sec heta–1}{ an heta–\sec heta+1}]$.
Use the trigonometric identity, $\sec^2 heta– an^2 heta=1$
⇒ $[\frac{ an heta+\sec heta–(\sec^2 heta– an^2 heta)}{ an heta–\sec heta+1}]=[\frac{( an heta+\sec heta)(1–\sec heta+ an heta)}{ an heta–\sec heta+1}$
$= an heta+\sec heta=\frac{1+\sin heta}{\cos heta}$
Hence, the correct answer is $\frac{1+\sin heta}{\cos heta}$.

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Question : Which of the following is equal to [tan⁡θ+sec⁡θ–1tan⁡θ–sec⁡θ+1]?Option 1: 1+sin⁡θcos⁡θOption 2: 1+tan⁡θcot⁡θOption 3: 1+cot⁡θtan⁡θOption 4: 1+cos⁡θsin⁡θ

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Question : Which of the following is equal to $[\frac{\tan θ + \sec θ - 1}{\tan θ - \sec θ + 1}]$ ?
Option 1: $\frac{1 + \sin θ}{\cos θ}$
Option 2: $\frac{1 + \tan θ}{\cot θ}$
Option 3: $\frac{1 + \cot θ}{\tan θ}$
Option 4: $\frac{1 + \cos θ}{\sin θ}$