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Miscellaneous Exercise 2 · Q59

Q.If sec⁡θ=m\sec\theta = m and tan⁡θ=n\tan\theta = n, then (m+n)+1m+n(m+n) + \dfrac{1}{m+n} is equal to: A) 2 B) mnmn C) 2m2m D) 2n2n

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Step 1. m=sec⁡θ,n=tan⁡θm=\sec\theta, n=\tan\theta, and sec⁡2θ−tan⁡2θ=1⇒(m−n)(m+n)=1⇒1m+n=m−n\sec^2\theta-\tan^2\theta=1 \Rightarrow (m-n)(m+n)=1 \Rightarrow \dfrac1{m+n}=m-n. …

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