COMEDK · Maths · 8. Trigonometric Ratios & Identities
If \(\sec \theta=m\) and \(\tan \theta=n\), then \(\frac{1}{m}\left[(m+n)+\frac{1}{(m+n)}\right]=\)
- A \(m n\)
- B \(2 n\)
- C \(2 m\)
- D 2
Answer & Solution
Correct Answer
(D) 2
Step-by-step Solution
Detailed explanation
We have, \(\sec \theta=m\) and \(\tan \theta=n\) Now, \(\frac{1}{m}\left[(m+n)+\frac{1}{(m+n)}\right]\) \(=\frac{1}{\sec \theta}\left[(\sec \theta+\tan \theta)+\frac{1}{(\sec \theta+\tan \theta)}\right]\)…
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